Derivative Exercise: 17.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

The Derivative 

Exercise 17.1

1. Find, From definition, the derivatives of the following:

(i) 3x²

Solution:

Let y = 3x²

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = 3(x + ∆x)²

Or, ∆y = 3(x + ∆x)² – y

Or, ∆y = 3(x + ∆x)² – 3x²

= 3{(x + ∆x)² – x²}

= 3{x² + 2x.∆x + (∆x)² – x²} = 3{2.x∆x + (∆x)²}

So, dy = 3∆x(2x + ∆x)

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{3\Delta {\rm{x}}\left( {2{\rm{x}} + \Delta {\rm{x}}} \right)}}{{\Delta {\rm{x}}}}$ = 3(2x + ∆x)

= ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 3(2x + ∆x)

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 3(2x + 0) = 6x.

 

(ii)x² – 2

Solution:

Let y = x² – 2

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = (x + ∆x)² – 2

Or, ∆y = (x + ∆x)² – y

 = (x + ∆x)² –2 – (x² – 2)

= x² + 2x. ∆x + (∆x)² – 2 – x² + 2

∆y = ∆x(2x + ∆x)

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{\Delta {\rm{x}}\left( {2{\rm{x}} + \Delta {\rm{x}}} \right)}}{{\Delta {\rm{x}}}}$ = 2x + ∆x

 ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta{\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 3(2x + ∆x)

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 2x + 0 = 2x.

 

(iii) x² + 5x – 3

Solution:

Let y = x² + 5x – 3

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = (x + ∆x)² + 5(x + ∆x) – 3

Or, ∆y = (x + ∆x)² + 5(x + ∆x) – 3 = (x + ∆x)² + 5(x + ∆x) – 3 – y

 = (x + ∆x)² + 5(x + ∆x) – 3 – (x² + 5x – 3)

= x² + 2x. ∆x + (∆x)² + 5x + 5∆x – 3 – x² – 5x + 3

= ∆x(2x + ∆x + 5)

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{\Delta {\rm{x}}\left( {2{\rm{x}} + \Delta {\rm{x}} + 5} \right)}}{{\Delta {\rm{x}}}}$ = 2x + ∆x + 5

 ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 3(2x + ∆x + 5)

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 6x + 0 + 5 = 2x + 5.

 

(iv) 3x² – 2x + 1

Solution:
Let y = 3x² – 2x + 1

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = 3(x + ∆x)² – 2(x + ∆x) + 1

Or, ∆y = 3x² + 6x. ∆x + 3(∆x)² – 2x – 2∆x + 1 – (3x² – 2x + 1)

 = 3x² + 6x∆x + 3(∆x)² – 2x – 2∆x + 1 – 3x² + 2x – 1.

= ∆x(6x + 3∆x – 2)

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = (6x + 3∆x – 2)

 ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 (6x + 3∆x + 2)

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 6x – 2.

 

(v)$\frac{1}{{\rm{x}}}$

Solution:

Let y = $\frac{1}{{\rm{x}}}$

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\frac{1}{{{\rm{x}} + \Delta {\rm{x\: }}}}$

Or, ∆y = $\frac{1}{{{\rm{x}} + \Delta {\rm{x\: }}}}$–y=$\frac{1}{{{\rm{x}} + \Delta {\rm{x\: }}}}$–$\frac{1}{{\rm{x}}}$ = $\frac{{{\rm{x}} - {\rm{x}} - \Delta {\rm{x\: }}}}{{{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}}$ = $\frac{{ - \Delta {\rm{x\: }}}}{{{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $ - \frac{1}{{{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}}$

 ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0$ - \frac{1}{{{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{1}{{{\rm{x}}\left( {{\rm{x}} + 0} \right)}}$ = $ - \frac{1}{{{{\rm{x}}^2}}}$.

 

(vi)$\frac{3}{{2{{\rm{x}}^2}}}$

Solution:

Let y = $\frac{3}{{2{{\rm{x}}^2}}}$

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\frac{3}{{2{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}{\rm{\: }}}}$

Or, ∆y = $\frac{3}{{2{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}{\rm{\: }}}}$– y = $\frac{3}{{2{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}{\rm{\: }}}}$–$\frac{3}{{2{{\rm{x}}^2}}}$ = $\frac{3}{2}\left[ {\frac{{{{\rm{x}}^2} - {{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}{\rm{\: }}}}{{{{\rm{x}}^2}{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}}} \right]$ = $\frac{3}{2}.\frac{{{{\rm{x}}^2} - {{\rm{x}}^2} - 2{\rm{x}}\Delta {\rm{x}} - {{\left( {\Delta {\rm{x}}} \right)}^2}.{\rm{\: }}}}{{{{\rm{x}}^2}{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}}$

= $ - \frac{{\frac{3}{2}\left( {\Delta {\rm{x}}\left( {2{\rm{x}} + \Delta {\rm{x}}} \right)} \right)}}{{{{\rm{x}}^2}{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}}}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $ - \frac{{\frac{3}{2}\left( {2{\rm{x}} + \Delta {\rm{x}}} \right)}}{{{{\rm{x}}^2}{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}}}$

 ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $ - \frac{{\frac{3}{2}\left( {2{\rm{x}} + \Delta {\rm{x}}} \right)}}{{{{\rm{x}}^2}{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{\frac{3}{2}\left( {2{\rm{x}}} \right)}}{{{{\rm{x}}^2}.{{\rm{x}}^2}}}$ = $ - \frac{3}{{{{\rm{x}}^3}}}$.

 

(vii)$\frac{1}{{{\rm{x}} - 1}}$

Solution:

Let y = $\frac{1}{{{\rm{x}} - 1}}$

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\frac{1}{{{\rm{x}} + \Delta {\rm{x}} - 1{\rm{\: }}}}$

Or, ∆y = $\frac{1}{{{\rm{x}} + \Delta {\rm{x}} - 1{\rm{\: }}}}$– y = $\frac{3}{{{\rm{x}} + \Delta {\rm{x}} - 1{\rm{\: }}}}$–$\frac{1}{{{\rm{x}} - 1}}$ = $\frac{{{\rm{x}} - 1 - {\rm{x}} - \Delta {\rm{x}} + 1}}{{\left( {{\rm{x}} + \Delta {\rm{x}} - 1} \right)\left( {{\rm{x}} - 1} \right)}}$ = $ - \frac{{\Delta {\rm{x}}}}{{\left( {{\rm{x}} + \Delta {\rm{x}} - 1} \right)\left( {{\rm{x}} - 1} \right)}}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $ - \frac{1}{{\left( {{\rm{x}} + \Delta {\rm{x}} - 1} \right)\left( {{\rm{x}} - 1} \right)}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{1}{{\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} - 1} \right)}}$ = $ - \frac{1}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}$.

 

(viii)$\frac{1}{{5{\rm{x}} - 1}}$

Solution:

Let y = $\frac{1}{{5{\rm{x}} - 1}}$

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\frac{1}{{5 - \left( {{\rm{x}} + \Delta {\rm{x}}} \right){\rm{\: }}}}$

Or, ∆y = $\frac{1}{{5 - {\rm{x}} - \Delta {\rm{x\: }}}}$– y = $\frac{1}{{5 - {\rm{x}} - \Delta {\rm{x\: }}}}$–$\frac{1}{{5 - {\rm{x}}}}$ = $\frac{{5 - {\rm{x}} - 5 + {\rm{x}} + \Delta {\rm{x}} + 1}}{{\left( {5 - {\rm{x}} - \Delta {\rm{x}}} \right)\left( {5 - {\rm{x}}} \right)}}$ = $\frac{{\Delta {\rm{x}}}}{{\left( {5 - {\rm{x}} - \Delta {\rm{x}}} \right)\left( {5 - {\rm{x}}} \right)}}$

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{\Delta {\rm{x}}}}{{\Delta {\rm{x}}\left( {5 - {\rm{x}} - \Delta {\rm{x}}} \right)\left( {5 - {\rm{x}}} \right)}}$ = $\frac{1}{{\left( {5 - {\rm{x}} - \Delta {\rm{x}}} \right)\left( {5 - {\rm{x}}} \right)}}$.

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{1}{{\left( {5 - {\rm{x}} - \Delta {\rm{x}}} \right)\left( {5 - {\rm{x}}} \right)}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{\left( {5 - {\rm{x}}} \right)\left( {5 - {\rm{x}}} \right)}}$ = $ - \frac{1}{{{{\left( {5 - {\rm{x}}} \right)}^2}}}$.

 

(ix)$\frac{1}{{2{\rm{x}} + 3}}$

Solution:

Let y = $\frac{1}{{2{\rm{x}} + 3}}$

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\frac{1}{{2\left( {{\rm{x}} + \Delta {\rm{x}}} \right) + 3{\rm{\: }}}}$

Or, ∆y = $\frac{1}{{2{\rm{x}} + 2\Delta {\rm{x}} + 3{\rm{\: }}}}$– y = $\frac{1}{{2{\rm{x}} + 2\Delta {\rm{x}} + 3{\rm{\: }}}}$–$\frac{1}{{2{\rm{x}} + 3}}$ = $\frac{{2{\rm{x}} + 3 - 2{\rm{x}} - 2\Delta {\rm{x}} - 3}}{{\left( {2{\rm{x}} + 2\Delta {\rm{x}} + 3} \right)\left( {2{\rm{x}} + 3} \right)}}$

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{{ - 2\Delta {\rm{x}}}}{{\left( {2{\rm{x}} + 2\Delta {\rm{x}} + 3} \right)\left( {2{\rm{x}} + 3} \right)}}$ = $ - \frac{2}{{\left( {2{\rm{x}} + 2\Delta {\rm{x}} + 3} \right)\left( {2{\rm{x}} + 3} \right)}}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{ - 2}}{{\left( {2{\rm{x}} + 2\Delta {\rm{x}} + 3} \right)\left( {2{\rm{x}} + 3} \right)}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{ - 2}}{{\left( {2{\rm{x}} + 2\Delta {\rm{x}} + 3} \right)\left( {2{\rm{x}} + 3} \right)}}$ = $ - \frac{2}{{{{\left( {2{\rm{x}} + 3} \right)}^2}}}$.

 

(x)$\frac{{{\rm{ax}}}}{{\rm{x}}} + \frac{{\rm{b}}}{{\rm{a}}} = {\rm{a}} + \frac{{\rm{b}}}{{\rm{x}}}$

Solution:

Let y = $\frac{{{\rm{ax}}}}{{\rm{x}}} + \frac{{\rm{b}}}{{\rm{a}}} = {\rm{a}} + \frac{{\rm{b}}}{{\rm{x}}}$

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = ${\rm{a}} + \frac{{\rm{b}}}{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right){\rm{\: }}}}$

Or, ∆y = ${\rm{a}} + \frac{{\rm{b}}}{{{\rm{x}} + \Delta {\rm{x\: }}}}$– y = ${\rm{a}} + \frac{{\rm{b}}}{{{\rm{x}} + \Delta {\rm{x\: }}}} - {\rm{a}} - \frac{{\rm{b}}}{{\rm{x}}}$

= ${\rm{b}}\left( {\frac{1}{{{\rm{x}} + \Delta {\rm{x}}}} - \frac{1}{{\rm{x}}}} \right)$ = b${\rm{\: }}\frac{{{\rm{x}} - {\rm{x}} - \Delta {\rm{x}}}}{{{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}$ = $ - \frac{{{\rm{b}}\Delta {\rm{x}}}}{{{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}$.

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $ - \frac{{{\rm{b}}\Delta {\rm{x}}}}{{\Delta {\rm{x}}.{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}$ = $ - \frac{{\rm{b}}}{{{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{ - {\rm{b}}}}{{{\rm{x}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{ - {\rm{b}}}}{{{\rm{x}}\left( {{\rm{x}} + 0} \right)}}$ = $ - \frac{{\rm{b}}}{{{{\rm{x}}^2}}}$.

 

(xi)x^1/2

Solution:

Let y = x^1/2 = $\sqrt {\rm{x}} $

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\sqrt {{\rm{x}} + \Delta {\rm{x}}} $

Or, ∆y = $\sqrt {{\rm{x}} + \Delta {\rm{x}}} $– y = $\sqrt {{\rm{x}} + \Delta {\rm{x}}}  - \sqrt {\rm{x}} $

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  - \sqrt {\rm{x}} }}{{\Delta {\rm{x}}}}$ = $\frac{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  - \sqrt {\rm{x}} }}{{\Delta {\rm{x}}}}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  - \sqrt {\rm{x}} }}{{\Delta {\rm{x}}}}$.

= ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  - \sqrt {\rm{x}} }}{{\Delta {\rm{x}}}}{\rm{*}}\frac{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} }}{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} }}$

= ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{\rm{x}} + \Delta {\rm{x}} - {\rm{x}}}}{{\Delta {\rm{x}}\left( {\sqrt {{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} } \right)}}$

= ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{x}}}}{{\Delta {\rm{x}}\left( {\sqrt {{\rm{x}} +\Delta {\rm{x}}}  + \sqrt {\rm{x}} } \right)}}$

= ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{1}{{\sqrt {{\rm{x}} +\Delta{\rm{x}}}  + \sqrt {\rm{x}} }}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{\sqrt {{\rm{x}} + 0}  + \sqrt {\rm{x}} }}$ = $\frac{1}{{2\sqrt {\rm{x}} }}$.

 

(xii)x + $\sqrt {\rm{x}} $

Solution:

Let y = x + $\sqrt {\rm{x}} $

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = x + ∆x + $\sqrt {{\rm{x}} + \Delta {\rm{x}}} $

Or, ∆y = x + ∆x + $\sqrt {{\rm{x}} + \Delta {\rm{x}}} $ – y.

= x + ∆x + $\sqrt {{\rm{x}} + \Delta {\rm{x}}} $– x –$\sqrt {\rm{x}} $ = ∆x + ($\sqrt {{\rm{x}} + \Delta {\rm{x}}}  - \sqrt {\rm{x}} $)

= ∆x + $\left( {\sqrt {{\rm{x}} + \Delta {\rm{x}}}  - \sqrt {\rm{x}} } \right)$ * $\frac{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} }}{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} }}$ = ∆x + $\frac{{{\rm{x}} + \Delta {\rm{x}} - {\rm{x}}}}{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} }}$.

∆y = ∆x + $\frac{{\Delta {\rm{x}}}}{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} }}$

Dividing both sides by ∆x.

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = 1 + $\frac{{\Delta {\rm{x}}}}{{\sqrt {{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} }}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left[ {1 + \frac{1}{{\sqrt{{\rm{x}} + \Delta {\rm{x}}}  + \sqrt {\rm{x}} }}} \right]{\rm{\: }}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $1 + \frac{1}{{\sqrt {\rm{x}}  + \sqrt {\rm{x}} }}$ = 1 + $\frac{1}{{2\sqrt {\rm{x}} }}$.

 

(xiii)(1 + x)^1/2

Solution:

Let y = (1 + x)^1/2 = $\sqrt {1 + {\rm{x}}} $

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\sqrt {1 + {\rm{x}} + \Delta {\rm{x}}} $

Or, ∆y = $\sqrt {1 + {\rm{x}} + \Delta {\rm{x}}} $ – y.

= $\sqrt {1 + {\rm{x}} + \Delta {\rm{x}}}  - \sqrt {1 + {\rm{x}}} $

= $\frac{{\left( {\sqrt {1 + {\rm{x}} + \Delta {\rm{x\: }}}  - \sqrt {1 + {\rm{x}}} } \right)\left( {\sqrt {1 + {\rm{x}} + \Delta {\rm{x\: }}}  - \sqrt {1 + {\rm{x}}} } \right)}}{{\left( {\sqrt {1 + {\rm{x}} + \Delta {\rm{x\: }}}  + \sqrt {1 + {\rm{x}}} } \right)}}$

∆y =$\frac{{1 + {\rm{x}} + \Delta {\rm{x}} - 1 - {\rm{x}}}}{{\sqrt {1 + {\rm{x}} + \Delta {\rm{x\: }}}  + \sqrt {1 + {\rm{x}}} }}$ = ∆x + $\frac{{\Delta {\rm{x}}}}{{\sqrt {1 + {\rm{x}} + \Delta {\rm{x}}}  + \sqrt {1 + {\rm{x}}} }}$

Dividing both sides by ∆x.

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{1}{{\sqrt {1 + {\rm{x}} + \Delta {\rm{x}}}  + \sqrt {1 + {\rm{x}}} }}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{1}{{\sqrt {1 + {\rm{x}} + \Delta {\rm{x}}}  + \sqrt {1 + {\rm{x}}} }}{\rm{\: }}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{\sqrt {1 + {\rm{x}} + 0}  + \sqrt {1 + {\rm{x}}} }}$ = $\frac{1}{{2\sqrt {1 + {\rm{x}}} }}$.

 

(xiv) (2x + 3)^1/2

Solution: 

Let y = (2x + 3)^1/2 = $\sqrt {2{\rm{x}} + 3} $

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\sqrt {2\left( {{\rm{x}} + \Delta {\rm{x}}} \right) + 3} $

Or, ∆y = $\sqrt {2\left( {{\rm{x}} + \Delta {\rm{x}}} \right) + 3} $ –$\sqrt {2{\rm{x}} + 3} $.

= $\frac{{\left( {\sqrt {2{\rm{x}} + 2\Delta {\rm{x}} + 3}  - \sqrt {2{\rm{x}} + 3} } \right)\left( {\sqrt {2{\rm{x}} + 2\Delta {\rm{x}} + 3}  + \sqrt {2{\rm{x}} + 3} } \right)}}{{\sqrt {2{\rm{x}} + 2\Delta {\rm{x}} + 3}  + \sqrt {2{\rm{x}} + 3} }}$

= $\frac{{2{\rm{x}} + 2\Delta {\rm{x}} + 3 - 2{\rm{x}} - 3}}{{\left( {\sqrt {2{\rm{x}} + 2\Delta  + 3{\rm{\: }}}  + \sqrt {2{\rm{x}} + 3} } \right)}}$

= $\frac{{2\Delta {\rm{x}}}}{{\sqrt {2{\rm{x}} + 2\Delta {\rm{x}} + 3}  + \sqrt {2{\rm{x}} + 3} }}$

Dividing both sides by ∆x.

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{2}{{\sqrt {2{\rm{x}} + 2\Delta {\rm{x}} + 3}  + \sqrt {2{\rm{x}} + 3} }}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{2}{{\sqrt {2{\rm{x}} + 2\Delta {\rm{x}} + 3}  + \sqrt {2{\rm{x}} + 3} }}{\rm{\: }}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{2}{{\sqrt {2{\rm{x}} + 0 + 3}  + \sqrt {2{\rm{x}} + 3} }}$ = $\frac{1}{{\sqrt {2{\rm{x}} + 3} }}$.

 

(xv)(1 + x²)^1/2

Solution:

Let y = (1 + x²)^1/2 = $\sqrt {1 + {{\rm{x}}^2}} $

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = $\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}} $

Or, ∆y = $\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}^2}} $ – $\sqrt {1 + {{\rm{x}}^2}} $.

= $\frac{{\left( {\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}  - \sqrt {1 + {{\rm{x}}^2}} } \right)\left( {\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}  + \sqrt {1 + {{\rm{x}}^2}} } \right)}}{{\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}  + \sqrt {1 + {{\rm{x}}^2}} }}$

= $\frac{{1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2} - 1 - {{\rm{x}}^2}}}{{\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}  + \sqrt {1 + {{\rm{x}}^2}} }}$ = $\frac{{1 + {{\rm{x}}^2} + 2{\rm{x}}.\Delta {\rm{x}} + {{\left( {\Delta {\rm{x\: }}} \right)}^2} - 1 - {{\rm{x}}^2}}}{{\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}  + \sqrt {1 + {{\rm{x}}^2}} }}$.

∆y = $\frac{{\Delta {\rm{x\: }}\left( {2{\rm{x}} + \Delta {\rm{x\: }}} \right)}}{{\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}  + \sqrt {1 + {{\rm{x}}^2}} }}$

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{{2{\rm{x}} + \Delta {\rm{x}}}}{{\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}  + \sqrt {1 + {{\rm{x}}^2}} }}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta{\rm{x}}}}$ = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{2{\rm{x}} + \Delta {\rm{x}}}}{{\sqrt {1 + {{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^2}}  + \sqrt {1 + {{\rm{x}}^2}} }}{\rm{\: }}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}} + 0}}{{\sqrt {1 + {{\left( {{\rm{x}} + 0{\rm{\: }}} \right)}^2}}  + \sqrt {1 + {{\rm{x}}^2}} }}$ = $\frac{{2{\rm{x}}}}{{2\sqrt {1 + {{\rm{x}}^2}} }}$ = $\frac{{\rm{x}}}{{\sqrt {1 + {{\rm{x}}^2}} }}$.

 

(xvi)$\frac{1}{{{{\rm{x}}^{\frac{1}{2}}}}}$= x^–1/2

Solution:

Let y =$\frac{1}{{{{\rm{x}}^{\frac{1}{2}}}}}$= x^–1/2

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = (x + ∆x)^–1/2

Or, ∆y = (x + ∆x)^–1/2 – x^–1/2.

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{{{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^{ - \frac{1}{2}}} - {{\rm{x}}^{ - \frac{1}{2}}}}}{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right) - {\rm{x}}}}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{1}{2}.{{\rm{x}}^{ - \frac{1}{2} - 1}}$$\left[ {{\rm{x\: }}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}{\rm{\: a}}\frac{{{{\rm{x}}^{\rm{n}}} - {{\rm{a}}^{\rm{n}}}}}{{{\rm{x}} - {\rm{a}}}} = {\rm{n}}{{\rm{a}}^{{\rm{n}} - 1}}} \right]{\rm{\: }}$

= $ - \frac{1}{2}{{\rm{x}}^{ - \frac{3}{2}}}$ = $ - \frac{1}{{2{{\rm{x}}^{\frac{3}{2}}}}}$.

 

(xvii)$\frac{1}{{{{\left( {1 - {\rm{x}}} \right)}^{\frac{1}{2}}}}}$ = (1 – x^–1/2)

Solution:

Let y = $\frac{1}{{{{\left( {1 - {\rm{x}}} \right)}^{\frac{1}{2}}}}}$ = (1 – x^–1/2)

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = {1 – (x + ∆x)}^–1/2

Or, ∆y = (1 – x – ∆x)^–1/2 – (1 – x)^–1/2.

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{{{{\left( {1 - {\rm{x}} - \Delta {\rm{x\: }}} \right)}^{ - \frac{1}{2}}} - {{\left( {1 - {\rm{x}}} \right)}^{ - \frac{1}{2}}}}}{{\Delta {\rm{x}}}}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$   = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\left( {1 - {\rm{x}} - \Delta {\rm{x\: }}} \right)}^{ - \frac{1}{2}}} - {{\left( {1 - {\rm{x}}} \right)}^{ - \frac{1}{2}}}}}{{\left( {1 - {\rm{x}} - \Delta {\rm{x\: }}} \right) - \left( {1 - {\rm{x}}} \right)}}{\rm{*}}\left( { - 1} \right)$

= (1 – x – ∆x) $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 (1 – x) $\frac{{{{\left( {1 - {\rm{x}} - \Delta {\rm{x\: }}} \right)}^{ - \frac{1}{2}}} - {{\left( {1 - {\rm{x}}} \right)}^{ - \frac{1}{2}}}}}{{\left( {1 - {\rm{x}} - \Delta {\rm{x\: }}} \right) - \left( {1 - {\rm{x}}} \right)}}{\rm{*}}\left( { - 1} \right)$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{1}{2}$(1 – x)^–1/2–1(–1) = $\frac{1}{{2{{\left( {1 - {\rm{x}}} \right)}^{\frac{3}{2}}}}}$.

 

(xviii)$\frac{1}{{{{\left( {3{\rm{x}} + 4} \right)}^{\frac{1}{2}}}}}$ = (3x + 4)^–1/2

Solution:

Let y = $\frac{1}{{{{\left( {3{\rm{x}} + 4} \right)}^{\frac{1}{2}}}}}$ = (3x + 4)^–1/2

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = {3(x + ∆x) + 4}^–1/2

Or, ∆y = (3x + 3∆x + 4)^–1/2 – (3x – 4)^–1/2.

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{{{{\left( {3{\rm{x}} + 3\Delta {\rm{x}} + 4{\rm{\: }}} \right)}^{ - \frac{1}{2}}} - {{\left( {3{\rm{x}} + 4} \right)}^{ - \frac{1}{2}}}}}{{\Delta {\rm{x}}}}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$   = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{{{\left( {3{\rm{x}} + 3\Delta {\rm{x}} + 4} \right)}^{ - \frac{1}{2}}} - {{\left( {3{\rm{x}} + 4} \right)}^{ - \frac{1}{2}}}}}{{\left( {3{\rm{x}} + 3\Delta {\rm{x}} + 4{\rm{\: }}} \right)--\left( {3{\rm{x}} + 4} \right)}}{\rm{*}}3$

= (3x + 3∆x + 4) $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 (3x + 4) $\frac{{{{\left( {3{\rm{x}} + 3\Delta {\rm{x}} + 4} \right)}^{ - \frac{1}{2}}} - {{\left( {3{\rm{x}} + 4} \right)}^{ - \frac{1}{2}}}}}{{\left( {3{\rm{x}} + 3\Delta {\rm{x}} + 4{\rm{\: }}} \right)--\left( {3{\rm{x}} + 4} \right)}}{\rm{*}}3$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{1}{2}$(3x + 4)^–1/2–1(3) = $\frac{3}{{2{{\left( {3{\rm{x}} + 4} \right)}^{\frac{3}{2}}}}}$.    

 

(xix)$\frac{{{\rm{ax}} + {\rm{b}}}}{{\sqrt {\rm{x}} }}$ = $\frac{{{\rm{ax}}}}{{\sqrt {\rm{x}} }}$ + $\frac{{\rm{b}}}{{\sqrt {\rm{x}} }}$

Solution:

Let y = $\frac{{{\rm{ax}} + {\rm{b}}}}{{\sqrt {\rm{x}} }}$ = $\frac{{{\rm{ax}}}}{{\sqrt {\rm{x}} }}$ + $\frac{{\rm{b}}}{{\sqrt {\rm{x}} }}$

y = ax^1/2 + bx^–1/2

Let ∆x and ∆y be the small increments in x and y respectively.

Or, y + ∆y = a(x + ∆x)^1/2 + b(x + ∆x)^–1/2

Or, ∆y = a(x + ∆x)^1/2 + b(x + ∆x)^–1/2 – ax^1/2 – bx^–1/2

= a{(x + ∆x)^1/2 – x^1/2} + b{(x + ∆x)^–1/2 – x^–1/2}.

$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$  = $\frac{{{\rm{a}}\left\{ {{{\left( {{\rm{x\: }} + {\rm{\: }}\Delta {\rm{x}}} \right)}^{\frac{1}{2}}}{\rm{\: }}--{\rm{\: }}{{\rm{x}}^{\frac{1}{2}}}} \right\}}}{{\Delta {\rm{x}}}}$ + $\frac{{{\rm{b}}\left\{ {\left( {{\rm{x\: }} + {\rm{\: }}\Delta {\rm{x}}} \right){ - ^{\frac{1}{2}}}{\rm{\: }}--{\rm{\: }}{{\rm{x}}^{\frac{{ - 1}}{2}}}} \right\}}}{{\Delta {\rm{x}}}}$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$   = ∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left[ {{\rm{a}}.\frac{{{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^{\frac{1}{2}}} - {{\rm{x}}^{\frac{1}{2}}}}}{{\Delta {\rm{x}}}} + {\rm{b}}.\frac{{{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^{ - \frac{1}{2}}} - {{\rm{x}}^{ - \frac{1}{2}}}}}{{\Delta {\rm{x}}}}} \right]$

∆x $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ 0 $\left[ {{\rm{a}}.\frac{{{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^{\frac{1}{2}}} - {{\rm{x}}^{\frac{1}{2}}}}}{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right) - {\rm{x}}}} + {\rm{b}}.\frac{{{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^{ - \frac{1}{2}}} - {{\rm{x}}^{ - \frac{1}{2}}}}}{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right) - {\rm{x}}}}} \right]$

(x + ∆x) $\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}$ x $\left[ {{\rm{a}}.\frac{{{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^{\frac{1}{2}}} - {{\rm{x}}^{\frac{1}{2}}}}}{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right) - {\rm{x}}}} + {\rm{b}}.\frac{{{{\left( {{\rm{x}} + \Delta {\rm{x\: }}} \right)}^{ - \frac{1}{2}}} - {{\rm{x}}^{ - \frac{1}{2}}}}}{{\left( {{\rm{x}} + \Delta {\rm{x}}} \right) - {\rm{x}}}}} \right]$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = ${\rm{a}}.\frac{1}{2}$. x^1/2–1 + b.$\left( { - \frac{1}{2}} \right)$x^–1/2–1.

= a.$\frac{1}{2}$x^–1/2 – b.$\frac{1}{2}$x^–3/2 = $\frac{1}{{2{{\rm{x}}^{\frac{1}{2}}}}} - \frac{{\rm{b}}}{{2{{\rm{x}}^{\frac{3}{2}}}}}$ = $\frac{{{\rm{ax}} - {\rm{b}}}}{{2{{\rm{x}}^{\frac{3}{2}}}}}$.

 

2.Find the derivatives of the following:

(i)x⁵

Solution:

Let y = x⁵

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x⁵) = 5.x^5–1 = 5x⁴.

 

(ii)5x⁷

Solution:

Let y = 5x⁷

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(5x⁷) = 5$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x⁷) = 5.7x^7–1 = 35x⁶.

 

(iii)3x² – 5x + 7

Solution:

Let y = 3x² – 5x + 7

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3x² – 5x + 7) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3x²) – $\frac{{\rm{d}}}{{{\rm{dx}}}}$(5x) + $\frac{{\rm{d}}}{{{\rm{dx}}}}$(7) = 3.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x²) – 5$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x) + 0.

= 3.2.x^2–1 – 5.1 + 0 = 6x + 5.

(iv) $\frac{{3{{\rm{x}}^3} + 2{\rm{x}} - 1}}{{2{{\rm{x}}^2}}}$ = $\frac{3}{2}{\rm{x}} + \frac{1}{{\rm{x}}} - \frac{1}{{2{{\rm{x}}^2}}}$

Soln:

Let y = $\frac{{3{{\rm{x}}^3} + 2{\rm{x}} - 1}}{{2{{\rm{x}}^2}}}$ = $\frac{3}{2}{\rm{x}} + \frac{1}{{\rm{x}}} - \frac{1}{{2{{\rm{x}}^2}}}$

Y = $\frac{3}{2}{\rm{x}} + {{\rm{x}}^{ - 1}} - \frac{1}{2}{{\rm{x}}^{ - 2}}$

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{3}{2}{\rm{x}} + {{\rm{x}}^{ - 1}} - \frac{1}{2}{{\rm{x}}^{ - 2}}} \right)$

= $\frac{3}{2}.\frac{{\rm{d}}}{{{\rm{dx}}}}$(x) – $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^–1) – $\frac{1}{2}\frac{{\rm{d}}}{{{\rm{dx}}}}$ (x^–2) = 3.1 + (–1)x^–1–1 – $\frac{1}{2}$(–2)x^–2–1

= $\frac{3}{2}$– x^–2 + x^–3 = $\frac{3}{2} - \frac{1}{{{{\rm{x}}^2}}} + \frac{1}{{{{\rm{x}}^3}}}$ = $\frac{{3{{\rm{x}}^3} - 2{\rm{x}} + 2}}{{2{{\rm{x}}^3}}}$

 

(v)2x^3/4 – 3x^1/2 – 5x^1/4

Solution:

Let y = 2x^3/4 – 3x^1/2 – 5x^1/4

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {2{{\rm{x}}^{\frac{3}{4}}}{\rm{\: }}--{\rm{\: }}3{{\rm{x}}^{\frac{1}{2}}}{\rm{\: }}--{\rm{\: }}5{{\rm{x}}^{\frac{1}{4}}}} \right)$

= 2$.\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^3/4) – 3.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^1/2) – 5.$\frac{{\rm{d}}}{{{\rm{dx}}}}$ (x^1/4) = 2.$\frac{3}{4}$x^3/4–1 – $\frac{3}{2}$x^–1/2 – $\frac{5}{4}$x^–3/4 = $\frac{3}{2}{{\rm{x}}^{ - \frac{1}{4}}} - \frac{3}{2}{{\rm{x}}^{ - \frac{1}{2}}} - \frac{5}{4}{{\rm{x}}^{ - \frac{3}{4}}}$.

= $\frac{{6{{\rm{x}}^{\frac{1}{2}}} - 6{{\rm{x}}^{\frac{1}{4}}} - 5}}{{4{{\rm{x}}^{\frac{3}{4}}}}}$.

 

(vi)$\frac{{2{\rm{x}} + 3{{\rm{x}}^{\frac{3}{4}}} + {{\rm{x}}^{\frac{1}{2}}} + 1}}{{{{\rm{x}}^{\frac{1}{4}}}}}$ = $\frac{{2{\rm{x}}}}{{{{\rm{x}}^{\frac{1}{4}}}}}$ + $\frac{{3{{\rm{x}}^{\frac{3}{4}}}}}{{{{\rm{x}}^{\frac{1}{4}}}}}$ + $\frac{{{{\rm{x}}^{\frac{1}{2}}}}}{{{{\rm{x}}^{\frac{1}{4}}}}}$ + $\frac{1}{{{{\rm{x}}^{\frac{1}{4}}}}}$

Solution:

Let y = $\frac{{2{\rm{x}} + 3{{\rm{x}}^{\frac{3}{4}}} + {{\rm{x}}^{\frac{1}{2}}} + 1}}{{{{\rm{x}}^{\frac{1}{4}}}}}$ = $\frac{{2{\rm{x}}}}{{{{\rm{x}}^{\frac{1}{4}}}}}$ + $\frac{{3{{\rm{x}}^{\frac{3}{4}}}}}{{{{\rm{x}}^{\frac{1}{4}}}}}$ + $\frac{{{{\rm{x}}^{\frac{1}{2}}}}}{{{{\rm{x}}^{\frac{1}{4}}}}}$ + $\frac{1}{{{{\rm{x}}^{\frac{1}{4}}}}}$

y = 2x^3/4 + 3x^1/2 + x^1/4 + x^–1/4

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {2{{\rm{x}}^{\frac{3}{4}}}{\rm{\: }}--{\rm{\: }}3{{\rm{x}}^{\frac{1}{2}}} + {{\rm{x}}^{\frac{{ - 1}}{4}}}} \right)$

= 2$.\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^3/4) + 3.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^1/2) + .$\frac{{\rm{d}}}{{{\rm{dx}}}}$ (x^1/4) + $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^–1/4) = 2.$\frac{3}{4}{\rm{\: }}$x^3/4–1+3.$\frac{1}{2}$x^1/2–1+$\frac{1}{4}$x^1/4–1 + $\left( { - \frac{1}{4}} \right)$x^–1/4–1

= $\frac{3}{2}{{\rm{x}}^{ - \frac{1}{4}}} + \frac{3}{2}{{\rm{x}}^{ - \frac{1}{2}}} + \frac{1}{4}{{\rm{x}}^{ - \frac{3}{4}}} - \frac{1}{4}{{\rm{x}}^{ - \frac{5}{4}}}$.

= $\frac{3}{{2{{\rm{x}}^{\frac{1}{4}}}}} + \frac{3}{{2{{\rm{x}}^{\frac{1}{2}}}}} + \frac{1}{{4{{\rm{x}}^{\frac{3}{4}}}}} - \frac{1}{{4{{\rm{x}}^{\frac{5}{4}}}}}$.

 

(vii)x^3/4(x^2/3 + x^1/3 + 1) = x^17/12 + x^13/12 + x^3/4

Solution:

Let y = x^3/4(x^2/3 + x^1/3 + 1) = x^17/12 + x^13/12 + x^3/4

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^17/12) + $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^13/12) + $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^3/4)

= $\frac{{17}}{{12}}$. x^{17/12 – 1} + $\frac{{13}}{{12}}$x^{13/12 – 1}  + $\frac{3}{4}$ x^{3/4– 1.}

= $\frac{{17}}{{12}}$. X^5/12 + $\frac{{13}}{{12}}$ x^1/12 + $\frac{3}{4}$ x^–1/4.

 

(viii)(x^1/2 + x^–1/2)²

Solution:

Let y = (x^1/2 + x^–1/2)²

= x + 2.x^1/2.x^–1/2 + x^–1.

= x + 2 + x^–1.

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x + 2 + x^–1)

= $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x) + $\frac{{\rm{d}}}{{\rm{x}}}$(2) + $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x^–1) = 1 + 0 + (–1)x^–1–1 = 1 – $\frac{1}{{{{\rm{x}}^2}}}{\rm{\: }}$.

 

3. Using the product Rule Calculate the derivative of: 

(i)3x²(2x – 1)

Solution:

Let y = 3x²(2x – 1)

Differentiating both sides w.r.t. ‘x’,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}${3x²(2x – 1)}

= 3x²$\frac{{\rm{d}}}{{{\rm{dx}}}}$(2x – 1) + (2x – 1) $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3x²)

= 3x².2.1 + (2x – 1).3.2x = 6x² + 12x² – 6x

= 18x² – 6x = 6x(3x – 1).

 

(ii)(2x² + 1)(3x² – 2)

Solution:

Let y = (2x² + 1)(3x² – 2)

Differentiating both sides w.r.t. ‘x’,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}${(2x² + 1)(3x² – 2)}

= (2x² + 1)$\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{\: }}$(3x² – 2) + (3x² – 2) $\frac{{\rm{d}}}{{{\rm{dx}}}}$(2x² + 1)

= (2x² + 1).3.2x + (3x² – 2).2.2.x = 12x³ + 6x + 12x³ – 8x

= 24x³ – 2x = 2x(12x² – 1).

 

(iii)(3x⁴ + 5)(4x⁵ – 3)

Solution:

Let y = (3x⁴ + 5)(4x⁵ – 3)

Differentiating both sides w.r.t. ‘x’,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}${(3x⁴ + 5)(4x⁵ – 3)

= (3x⁴ + 5)$\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{\: }}$(4x⁵ – 3) + (4x⁵ – 3) $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3x⁴ + 5)

= (3x⁴ + 5).4.5.x⁴ + (4x⁵ – 3).3.4.x³

= 60x⁹ + 100x⁴ + 48x⁸ – 36x³ = 108x + 100x⁴ – 35x³

= 4x³(27x⁵ + 25x – 9)

 

(iv) (3x² + 5x – 1)(x² + 3)

Solution:

Let y = (3x² + 5x – 1)(x² + 3)

Differentiating both sides w.r.t. ‘x’,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}${(3x² + 5x – 1)( x² + 3)}

= (3x² + 5x – 1)$\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{\: }}$(x² + 3) + (x² + 3) $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3x² + 5x – 1)

= (3x² + 5x – 1).2x + (x² + 3).(3.2x + 5)

= 6x³ + 10x² – 2x + 6x³ + 18x + 5x² + 15

= 12x³ + 15x² + 16x + 15.

 

(v)(a + $\sqrt {\rm{x}} $)(a –$\sqrt {\rm{x}} $)

Solution:

Let y = (a + $\sqrt {\rm{x}} $)(a –$\sqrt {\rm{x}} $)

Differentiating both sides w.r.t. ‘x’,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =$\frac{{{\rm{d}}}}{{{\rm{dx}}}}${( a + $\sqrt {\rm{x}} $)( a – $\sqrt {\rm{x}} $)}

= (a + $\sqrt {\rm{x}} $)$\frac{{\rm{d}}}{{{{\rm{d}}^2}}}{\rm{\: }}$(a – $\sqrt {\rm{x}} $) + (a – $\sqrt {\rm{x}} $) $\frac{{\rm{d}}}{{{\rm{dx}}}}$(a + $\sqrt {\rm{x}} $)

= (a + $\sqrt {\rm{x}} $)$\left( { - \frac{1}{{2\sqrt {\rm{x}} }}} \right)$+ (a – $\sqrt {\rm{x}} $).$\frac{1}{{2\sqrt {\rm{x}} }}$

= $ - \frac{{\rm{a}}}{{2\sqrt {\rm{x}} }} - \frac{1}{2} + \frac{{\rm{a}}}{{2\sqrt {\rm{x}} }} - \frac{1}{2}$ = –1

 

(vi)(a + x^3/4)( a – x^1/4)

Solution:

Let y = (a + x^3/4)( a – x^1/4)

Differentiating both sides w.r.t. ‘x’,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}${(a + x^3/4)(a – x^1/4)}

= (a + x^3/4)$\left( { - \frac{1}{4}{{\rm{x}}^{ - \frac{3}{4}}}} \right)$+ (a – x^1/4).$\frac{3}{4}{{\rm{x}}^{ - \frac{1}{4}}}$

= $ - \frac{1}{4}{\rm{a}}{{\rm{x}}^{ - \frac{3}{4}}} - \frac{1}{4} + \frac{3}{4}{\rm{a}}{{\rm{x}}^{ - \frac{1}{4}}} - \frac{3}{4}$

= $\frac{{3{\rm{a}}}}{4}{{\rm{a}}^{ - \frac{1}{4}}} - \frac{1}{4}{\rm{a}}{{\rm{x}}^{ - \frac{3}{4}}} - 1$.

 

4.Using the quotient rule to find the derivative of:

(i)$\frac{{\rm{x}}}{{1 + {\rm{x}}}}$

Solution:

Let y = $\frac{{\rm{x}}}{{1 + {\rm{x}}}}$

Differentiating both sides w.r.t. x,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{\rm{x}}}{{1 + {\rm{x}}}}} \right)$.

= $\frac{{\left( {1 + {\rm{x}}} \right).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right) - {\rm{x}}.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {1 + {\rm{x}}} \right)}}{{{{\left( {1 + {\rm{x}}} \right)}^2}}}$ = $\frac{1}{{{{\left( {1 + {\rm{x}}} \right)}^2}}}$.

 

(ii)$\frac{{{{\rm{x}}^2}}}{{1 - {{\rm{x}}^2}}}$

Solution:

Let y = $\frac{{{{\rm{x}}^2}}}{{1 - {{\rm{x}}^2}}}$

Differentiating both sides w.r.t. x,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{{\rm{x}}^2}}}{{1 - {{\rm{x}}^2}}}} \right)$.

= $\frac{{\left( {1 - {{\rm{x}}^2}} \right).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right) - {{\rm{x}}^2}.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {1 - {{\rm{x}}^2}} \right)}}{{{{\left( {1 - {{\rm{x}}^2}} \right)}^2}}}$ = $\frac{{\left( {1 - {{\rm{x}}^2}} \right).2{\rm{x}} - {{\rm{x}}^2}.\left( { - 2{\rm{x}}} \right)}}{{{{\left( {1 - {{\rm{x}}^2}} \right)}^2}}}$.

= $\frac{{2{\rm{x}} - 2{{\rm{x}}^3} + 2{{\rm{x}}^3}}}{{{{\left( {1 - {{\rm{x}}^2}} \right)}^2}}}$ = $\frac{{2{\rm{x}}}}{{{{\left( {1 - {{\rm{x}}^2}} \right)}^2}}}$.

 

(iii)$\frac{{{{\rm{x}}^2} - {{\rm{a}}^2}}}{{{{\rm{x}}^2} + {{\rm{a}}^2}}}$

Solution:

Let y = $\frac{{{{\rm{x}}^2} - {{\rm{a}}^2}}}{{{{\rm{x}}^2} + {{\rm{a}}^2}}}$

Differentiating both sides w.r.t. x,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{{\rm{x}}^2} - {{\rm{a}}^2}}}{{{{\rm{x}}^2} + {{\rm{a}}^2}}}} \right)$.

= $\frac{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right) - ({{\rm{x}}^2} - {{\rm{a}}^2}).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}}{{{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}^2}}}$ = $\frac{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right).2{\rm{x}} - ({{\rm{x}}^2} - {{\rm{a}}^2}).2{\rm{x}}}}{{{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}^2}}}$.

=  $\frac{{4{{\rm{a}}^2}{\rm{x}}}}{{{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}^2}}}$

 

(iv) $\frac{3}{{{{\rm{x}}^2}}}$

Solution:

Let y = $\frac{3}{{{{\rm{x}}^2}}}$

Differentiating both sides w.r.t. x,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{3}{{{{\rm{x}}^2}}}} \right)$.

= $\frac{{{{\rm{x}}^2}.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( 3 \right) - 3.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right)}}{{{{\left( {{{\rm{x}}^2}} \right)}^2}}}$ = $\frac{{{{\rm{x}}^2}.0 - 3.2{\rm{x}}}}{{{{\rm{x}}^4}}}$. = $ - \frac{{6{\rm{x}}}}{{{{\rm{x}}^4}}}$ = $ - \frac{6}{{{{\rm{x}}^3}}}$.

 

(v)$\frac{{{{\rm{x}}^2} - 2{\rm{x}}}}{{{\rm{x}} + 1}}$

Solution:

Let y = $\frac{{{{\rm{x}}^2} - 2{\rm{x}}}}{{{\rm{x}} + 1}}$

Differentiating both sides w.r.t. x,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{{\rm{x}}^2} - 2{\rm{x}}}}{{{\rm{x}} + 1}}} \right)$.

= $\frac{{\left( {{\rm{x}} + 1} \right).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2} - 2{\rm{x}}} \right) - ({{\rm{x}}^2} - 2{\rm{x}}).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}} + 1} \right)}}{{{{\left( {{\rm{x}} + 1} \right)}^2}}}$ = $\frac{{\left( {{\rm{x}} + 1} \right).\left( {2{\rm{x}} - 2} \right) - ({{\rm{x}}^2} - 2{\rm{x}}).1}}{{{{\left( {{\rm{x}} + 1} \right)}^2}}}$ = $\frac{{2{{\rm{x}}^2} + 2{\rm{x}} - 2{\rm{x}} - 2 - {{\rm{x}}^2} + 2{\rm{x}}}}{{{{\left( {{\rm{x}} + 1} \right)}^2}}}$.

=  $\frac{{{{\rm{x}}^2} + 2{\rm{x}} - 2}}{{{{\left( {{\rm{x}} + 1} \right)}^2}}}$.

 

(vi)$\frac{{{{\rm{x}}^2} + 2{\rm{x}} - 1}}{{{{\rm{x}}^2} - 2{\rm{x}} + 1}}$

Solution:

Let y = $\frac{{{{\rm{x}}^2} + 2{\rm{x}} - 1}}{{{{\rm{x}}^2} - 2{\rm{x}} + 1}}$

Differentiating both sides w.r.t. x,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{{\rm{x}}^2} + 2{\rm{x}} - 1}}{{{{\rm{x}}^2} - 2{\rm{x}} + 1}}} \right)$.

= $\frac{{\left( {{{\rm{x}}^2} - 2{\rm{x}} + 1} \right).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2} + 2{\rm{x}} - 1} \right) - ({{\rm{x}}^2} + 2{\rm{x}} - 1).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2} - 2{\rm{x}} + 1} \right)}}{{{{\left( {{{\rm{x}}^2} - 2{\rm{x}} + 1} \right)}^2}}}$ = $\frac{{\left( {{{\rm{x}}^2} - 2{\rm{x}} + 1} \right).\left( {2{\rm{x}} + 2} \right) - ({{\rm{x}}^2} - 2{\rm{x}} + 1).\left( {2{\rm{x}} - 2} \right)}}{{{{\left( {{{\rm{x}}^2} - 2{\rm{x}} + 1} \right)}^2}}}$

= $\frac{{{{\left( {{\rm{x}} - 1} \right)}^2}.2\left( {{\rm{x}} + 1} \right) - \left( {{{\rm{x}}^2} + 2{\rm{x}} - 1} \right).2\left( {{\rm{x}} - 1} \right)}}{{{{\left\{ {{{\left( {{\rm{x}} + 1} \right)}^2}} \right\}}^2}}}$

= $\frac{{2\left( {{\rm{x}} - 1} \right)\left\{ {\left( {{\rm{x}} - 1} \right)\left( {{\rm{x}} + 1} \right) - \left( {{{\rm{x}}^2} + 2{\rm{x}} - 1} \right)} \right\}}}{{{{\left( {{\rm{x}} - 1} \right)}^4}}}{\rm{\: \: \: }}$

= $\frac{{2\left\{ {{{\rm{x}}^2} - 1 - {{\rm{x}}^2} - 2{\rm{x}} + 1} \right\}}}{{{{\left( {{\rm{x}} - 1} \right)}^3}}}$

= $ - \frac{{4{\rm{x}}}}{{{{\left( {{\rm{x}} - 1} \right)}^3}}}$

= $\frac{{4{\rm{x}}}}{{{{\left( {1 - {\rm{x}}} \right)}^3}}}$

 

(vii) $\frac{{\sqrt {\rm{x}} }}{{\sqrt {\rm{x}}  + 1}}$

Solution:

Let y = $\frac{{\sqrt {\rm{x}} }}{{\sqrt {\rm{x}}  + 1}}$

Differentiating both sides w.r.t. x,

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{\sqrt {\rm{x}} }}{{\sqrt {\rm{x}}  + 1}}} \right)$.

= $\frac{{\left( {\sqrt {\rm{x}}  + 1} \right).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\sqrt {\rm{x}} } \right) - \left( {\sqrt {\rm{x}} } \right).\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\sqrt {\rm{x}}  + 1} \right)}}{{{{\left( {\sqrt {\rm{x}}  + 1} \right)}^2}}}$ = $\frac{{\left( {\sqrt {\rm{x}}  + 1} \right).\frac{1}{{2\sqrt {\rm{x}} }} - \sqrt {\rm{x}} .\frac{1}{{2\sqrt {\rm{x}} }}}}{{{{\left( {\sqrt {\rm{x}}  + 1} \right)}^2}}}$

= $\frac{{\frac{{\sqrt {\rm{x}}  + 1 - \sqrt {\rm{x}} }}{{2\sqrt {\rm{x}} }}}}{{{{\left( {\sqrt {\rm{x}}  + 1} \right)}^2}}}$

= $\frac{1}{{2\sqrt {\rm{x}} {{\left( {\sqrt {\rm{x}}  + 1} \right)}^2}}}$.

 

5. Use the general; power rule to calculate the derivatives of:

(i)(2x + 3)²

Solution:

Let y = (2x + 3)²

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(2x + 3)²

= $\frac{{{\rm{d}}{{\left( {2{\rm{x}} + 3} \right)}^2}}}{{{\rm{d}}\left( {2{\rm{x}} + 3} \right)}}$.$\frac{{{\rm{d}}\left( {2{\rm{x}} + 3} \right)}}{{{\rm{dx}}}}$ = 2.(2x + 3).2.1 = 4(2x + 3)

 

(ii)(3 – 2x)²

Solution:

Let y = (3 – 2x)²

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3 – 2x)²

= $\frac{{{\rm{d}}{{\left( {3{\rm{\: }}--{\rm{\: }}2{\rm{x}}} \right)}^3}}}{{{\rm{d}}\left( {3 - 2{\rm{x}}} \right)}}$.$\frac{{{\rm{d}}\left( {3 - 2{\rm{x}}} \right)}}{{{\rm{dx}}}}$ = 3.(3 – 2x)².(–2.1) = –6(3 – 2x)².

 

(iii)(3x² + 2x – 1)⁴

Solution:

Let y = (3x² + 2x – 1)⁴

Differentiating both sides w.r.t. to x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{3{\rm{x}}}}$ (3x² + 2x – 1)⁴

= $\frac{{{\rm{d}}{{\left( {3{{\rm{x}}^2} + 2{\rm{x}} - 1} \right)}^4}}}{{{\rm{d}}\left( {3{{\rm{x}}^2} + 2{\rm{x}} - 1} \right)}}.\frac{{{\rm{d}}\left( {3{{\rm{x}}^2} + 2{\rm{x}} - 1} \right)}}{{{\rm{dx}}}}$

= 4.(3x² + 2x – 1)³.(6x + 2) = 4.(3x² + 2x – 1)³.2(3x + 1)

= 8(3x + 1)(3x² + 2x – 1)³.

 

(iv)(2x² + 3x – 3)^–6.

Solution:

Let y = (2x² + 3x – 3)^–6.

Differentiating both sides w.r.t. to x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(2x² + 3x – 3)^–6

= $\frac{{{\rm{d}}{{\left( {2{{\rm{x}}^2} + 3{\rm{x}} - 3} \right)}^{ - 6}}}}{{{\rm{d}}\left( {2{{\rm{x}}^2} + 3{\rm{x}} - 3} \right)}}.\frac{{{\rm{d}}\left( {2{{\rm{x}}^2} + 3{\rm{x}} - 3} \right)}}{{{\rm{dx}}}}$ = –6(2x² + 3x – 3)^–7(4x + 3) = –6(4x + 3)(2x² + 3x – 3)^–7.

 

(v)$\sqrt {8 - 5{\rm{x}}} $ = (8 – 5x)^1/2

Solution:

Let y = $\sqrt {8 - 5{\rm{x}}} $ = (8 – 5x)^1/2

Differentiating both sides w.r.t. x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(8 – 5x)^1/2

= $\frac{{{\rm{d}}{{\left( {8 - 5{\rm{x}}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {8 - 5{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {8 - 5{\rm{x}}} \right)}}{{{\rm{dx}}}}$ = $\frac{1}{2}.{\left( {8 - 5{\rm{x}}} \right)^{ - \frac{1}{2}}}\left( { - 5} \right)$ = $ - \frac{5}{{2\sqrt {8 - 5{\rm{x}}} }}$.

 

(vi) (2x² – 3x + 1)^3/4

Solution:

Let y = (2x² – 3x + 1)^3/4

Differentiating both sides w.r.t. to x.

= $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(2x² – 3x + 1)^3/3

= $\frac{{{\rm{d}}{{\left( {2{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)}^{\frac{3}{4}}}}}{{{\rm{d}}\left( {2{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)}}.\frac{{{\rm{d}}\left( {2{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)}}{{{\rm{dx}}}}$

= $\frac{3}{4}{\left( {2{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)^{ - \frac{1}{4}}}$(4x – 3) = $\frac{3}{4}$(4x – 3)(2x² – 3x + 1)^–1/4.

 

(vii)$\frac{1}{{\sqrt {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} }}$ = (ax² + bx + c)^–1/2

Solution:

Let y = $\frac{1}{{\sqrt {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} }}$ = (ax² + bx + c)^–1/2

Differentiating both sides w.r.t. to x.

= $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}$(ax² + bx + c)^–1/2

= $\frac{{{\rm{d}}{{\left( {{\rm{a}}{{\rm{x}}^2}{\rm{\: }} + {\rm{\: bx\: }} + {\rm{\: c}}} \right)}^{ - \frac{1}{2}}}{\rm{\: }}}}{{{\rm{d}}\left( {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} \right)}}{{{\rm{dx}}}}$

= $ - \frac{1}{2}{\left( {{\rm{a}}{{\rm{x}}^2} + {\rm{bx}} + {\rm{c}}} \right)^{ - \frac{3}{2}}}$.(a.2x + b.1)

= $ - \frac{1}{2}$(2ax + b)(ax² + bx + c)^–3/2.

 

(viii) $\frac{1}{{\sqrt[3]{{3{{\rm{x}}^2} - 4{\rm{x}} - 1}}}}$ = (3x² – 4x – 1)^–1/3

Solution:

Let y = $\frac{1}{{\sqrt[3]{{3{{\rm{x}}^2} - 4{\rm{x}} - 1}}}}$ = (3x² – 4x – 1)^–1/3

Differentiating both sides w.r.t. to x.

= $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}$(3x² – 4x – 1)^–1/3

= $\frac{{{\rm{d}}{{\left( {3{{\rm{x}}^2}{\rm{\: }}--{\rm{\: }}4{\rm{x\: }}--{\rm{\: }}1} \right)}^{ - \frac{1}{3}}}{\rm{\: }}}}{{{\rm{d}}\left( {3{{\rm{x}}^2}{\rm{\: }}--{\rm{\: }}4{\rm{x\: }}--{\rm{\: }}1} \right)}}.\frac{{{\rm{d}}\left( {3{{\rm{x}}^2}{\rm{\: }}--{\rm{\: }}4{\rm{x\: }}--{\rm{\: }}1} \right)}}{{{\rm{dx}}}}$

= $ - \frac{1}{3}{\left( {3{{\rm{x}}^2}{\rm{\: }}--{\rm{\: }}4{\rm{x\: }}--{\rm{\: }}1} \right)^{ - \frac{4}{3}}}$.(6x – 4)

= $ - \frac{1}{3}{\left( {3{{\rm{x}}^2}{\rm{\: }}--{\rm{\: }}4{\rm{x\: }}--{\rm{\: }}1} \right)^{ - \frac{4}{3}}}$(6x – 4)

= $ - \frac{1}{3}{\left( {3{{\rm{x}}^2}{\rm{\: }}--{\rm{\: }}4{\rm{x\: }}--{\rm{\: }}1} \right)^{ - \frac{4}{3}}}$2(3x – 2)

= $ - \frac{2}{3}$(3x – 2)(3x² – 4x – 1)^–4/3.

 

(ix) $\frac{1}{{\sqrt {{{\rm{a}}^{\rm{n}}} - {{\rm{x}}^{\rm{n}}}} }}$ = (aⁿ – xⁿ)^–1/2

Solution:

Let y = $\frac{1}{{\sqrt {{{\rm{a}}^{\rm{n}}} - {{\rm{x}}^{\rm{n}}}} }}$ = (aⁿ – xⁿ)^–1/2

Differentiating both sides w.r.t. to x.

= $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}$(aⁿ – xⁿ)^–1/2

= $\frac{{{\rm{d}}{{\left( {{{\rm{a}}^{\rm{n}}}{\rm{\: }}--{\rm{\: }}{{\rm{x}}^{\rm{n}}}} \right)}^{ - \frac{1}{2}}}{\rm{\: }}}}{{{\rm{d}}\left( {{{\rm{a}}^{\rm{n}}}{\rm{\: }}--{\rm{\: }}{{\rm{x}}^{\rm{n}}}} \right)}}.\frac{{{\rm{d}}\left( {{{\rm{a}}^{\rm{n}}} - {{\rm{x}}^{\rm{n}}}} \right)}}{{{\rm{dx}}}}$

= $ - \frac{1}{2}{\left( {{{\rm{a}}^{\rm{n}}} - {{\rm{x}}^{\rm{n}}}} \right)^{ - \left( {\frac{3}{2}} \right)}}.\left( { - {\rm{n}}{{\rm{x}}^{{\rm{n}} - 1}}} \right)$

= $\frac{1}{2}$nx^n–1(aⁿ – xⁿ)^–3/2.

 

(x) $\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}}  - \sqrt {\rm{x}} }}$ = $\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}}  - \sqrt {\rm{x}} }}{\rm{*}}\frac{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {\rm{x}} }}{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {\rm{x}} }}$

Solution:

Let y = $\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}}  - \sqrt {\rm{x}} }}$ = $\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}}  - \sqrt {\rm{x}} }}{\rm{*}}\frac{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {\rm{x}} }}{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {\rm{x}} }}$

= $\frac{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {\rm{x}} }}{{{\rm{x}} + {\rm{a}} - {\rm{x}}}}$ = $\frac{1}{{\rm{a}}}${(x + a)^1/2 + x^1/2}

Differentiating both sides w.r.t to x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{\rm{a}}}.\frac{{\rm{d}}}{{{\rm{dx}}}}${(x + a)^1/2 + x^1/2}

= $\frac{1}{{\rm{a}}}\left\{ {\frac{{{\rm{d}}{{\left( {{\rm{x}} + {\rm{a}}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{\rm{x}} + {\rm{a}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{x}} + {\rm{a}}} \right)}}{{{\rm{dx}}}} + \frac{{{\rm{d}}{{\rm{x}}^{\frac{1}{2}}}}}{{{\rm{dx}}}}} \right\}$

= $\frac{1}{{\rm{a}}}\left\{ {\frac{1}{2}{{\left( {{\rm{x}} + {\rm{a}}} \right)}^{ - \frac{1}{2}}}.1 + \frac{1}{2}{{\rm{x}}^{ - \frac{1}{2}}}} \right\}$ = $\frac{1}{{2{\rm{a}}}}\left( {\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}} }} + \frac{1}{{\sqrt {\rm{x}} }}} \right)$.

 

(xi)$\frac{1}{{{\rm{x}} - \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}$ = $\frac{1}{{{\rm{x}} - \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}{\rm{*}}\frac{{{\rm{x}} + \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}{{{\rm{x}} + \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}$

Solution:

Let y = $\frac{1}{{{\rm{x}} - \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}$ = $\frac{1}{{{\rm{x}} - \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}{\rm{*}}\frac{{{\rm{x}} + \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}{{{\rm{x}} + \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}$

= $\frac{{{\rm{x}} + \sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}{{{{\rm{x}}^2} - {{\rm{a}}^2} - {{\rm{x}}^2}}}$ = $ - \frac{1}{{{{\rm{a}}^2}}}${x + $\sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} $}

Differentiating both sides w.r.t to x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{1}{{{{\rm{a}}^2}}}.\frac{{\rm{d}}}{{{\rm{dx}}}}${x + (a² + x²)^1/2 }

= $\frac{1}{{{{\rm{a}}^2}}}\left\{ {\frac{{{\rm{dx}}}}{{{\rm{dx}}}} + \frac{{{\rm{d}}{{\left( {{{\rm{a}}^2} + {{\rm{x}}^2}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{{\rm{a}}^2} + {{\rm{x}}^2}} \right)}}.\frac{{{\rm{d}}\left( {{{\rm{a}}^2} + {{\rm{x}}^2}} \right)}}{{{\rm{dx}}}}} \right\}$

= $ - \frac{1}{{{{\rm{a}}^2}}}\left\{ {1 + \frac{1}{2}{{\left( {{{\rm{a}}^2} + {{\rm{x}}^2}} \right)}^{ - \frac{1}{2}}}.2{\rm{x}}} \right\}$ = $ - \frac{1}{{{{\rm{a}}^2}}}\left( {1 + \frac{1}{{\sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}} \right)$.

 

(xii)$\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {{\rm{x}} - {\rm{a}}} }}$ = $\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {{\rm{x}} - {\rm{a}}} }}{\rm{*}}\frac{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {{\rm{x}} - {\rm{a}}} }}{{\sqrt {{\rm{x}} + {\rm{a}}}  - \sqrt {{\rm{x}} - {\rm{a}}} }}$

Solution:

Let y = $\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {{\rm{x}} - {\rm{a}}} }}$ = $\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {{\rm{x}} - {\rm{a}}} }}{\rm{*}}\frac{{\sqrt {{\rm{x}} + {\rm{a}}}  + \sqrt {{\rm{x}} - {\rm{a}}} }}{{\sqrt {{\rm{x}} + {\rm{a}}}  - \sqrt {{\rm{x}} - {\rm{a}}} }}$

= $\frac{{\sqrt {{\rm{x}} + {\rm{a}}}  - \sqrt {{\rm{x}} - {\rm{a}}} }}{{{\rm{x}} + {\rm{a}} - {\rm{x}} + {\rm{a}}}}$ = $\frac{1}{{2{\rm{a}}}}\left\{ {{{\left( {{\rm{x}} + {\rm{a}}} \right)}^{\frac{1}{2}}} - {{\left( {{\rm{x}} - {\rm{a}}} \right)}^{\frac{1}{2}}}} \right\}$

Differentiating both sides w.r.t to x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{2{\rm{a}}}}.\frac{{\rm{d}}}{{{\rm{dx}}}}${(x + a)^1/2– (x – a)^1/2}

= $\frac{1}{{2{\rm{a}}}}\left\{ {\frac{{{\rm{d}}{{\left( {{\rm{x}} + {\rm{a}}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{\rm{x}} + {\rm{a}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{x}} + {\rm{a}}} \right)}}{{{\rm{dx}}}} - \frac{{{\rm{d}}{{\left( {{\rm{x}} - {\rm{a}}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{\rm{x}} - {\rm{a}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{x}} - {\rm{a}}} \right)}}{{{\rm{dx}}}}} \right\}$

= $\frac{1}{{2{\rm{a}}}}\left\{ {\frac{1}{2}{{\left( {{\rm{x}} + {\rm{a}}} \right)}^{ - \frac{1}{2}}}.1 - \frac{1}{2}{{\left( {{\rm{x}} - {\rm{a}}} \right)}^{ - \frac{1}{2}}}.1} \right\}$ = $\frac{1}{{4{\rm{a}}}}\left( {\frac{1}{{\sqrt {{\rm{x}} + {\rm{a}}} }} - \frac{1}{{\sqrt {{\rm{x}} - {\rm{a}}} }}} \right)$.

 

(xiii)$\frac{1}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  - \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$ = $\frac{1}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  - \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{\rm{*}}\frac{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  + \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  + \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$

Solution:

Let y = $\frac{1}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  - \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$ = $\frac{1}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  - \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{\rm{*}}\frac{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  + \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  + \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$

= $\frac{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}}  + \sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{{{{\rm{x}}^2} + {{\rm{a}}^2} - {{\rm{x}}^2} + {{\rm{a}}^2}}}$ = $\frac{1}{{2{{\rm{a}}^2}}}\left\{ {{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}^{\frac{1}{2}}} + {{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{\frac{1}{2}}}} \right\}$

Differentiating both sides w.r.t to x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{2{\rm{a}}}}.\frac{{\rm{d}}}{{{\rm{dx}}}}${(x + a)^1/2 – (x – a)^1/2}

= $\frac{1}{{2{{\rm{a}}^2}}}\left\{ {\frac{{{\rm{d}}{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}}.\frac{{{\rm{d}}\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}}{{{\rm{dx}}}} - \frac{{{\rm{d}}{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}}.\frac{{{\rm{d}}\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}}{{{\rm{dx}}}}} \right\}$

= $\frac{1}{{2{{\rm{a}}^2}}}\left\{ {\frac{1}{2}{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}^{ - \frac{1}{2}}}.2{\rm{x}} + \frac{1}{2}{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{ - \frac{1}{2}}}.2{\rm{x}}} \right\}$ = $\frac{{\rm{x}}}{{2{{\rm{a}}^2}}}\left( {\frac{1}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} }} + \frac{1}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}} \right)$.

 

(xiv)$\sqrt {\frac{{{{\rm{x}}^2} + {{\rm{a}}^2}}}{{{{\rm{x}}^2} - {{\rm{a}}^2}}}} $

Solution:

Let y = $\sqrt {\frac{{{{\rm{x}}^2} + {{\rm{a}}^2}}}{{{{\rm{x}}^2} - {{\rm{a}}^2}}}} $

Differentiating both sides w.r.t to x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} \frac{{\rm{d}}}{{{\rm{dx}}}}\sqrt {\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}  - \sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} \frac{{\rm{d}}}{{{\rm{dx}}}}\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{{{{\left( {\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} } \right)}^2}}}$.${\rm{\: }}$

= $\frac{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} .\frac{{{\rm{d}}{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}}.\frac{{{\rm{d}}\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}}{{{\rm{dx}}}} - \sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} .\frac{{{\rm{d}}{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}}.\frac{{{\rm{d}}\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}}{{{\rm{dx}}}}}}{{{{\rm{x}}^2} - {{\rm{a}}^2}}}$

= $\frac{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} .\frac{1}{2}{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}^{ - \frac{1}{2}}}.2{\rm{x}} - \sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} .\frac{1}{2}{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{ - \frac{1}{2}}}.2{\rm{x}}}}{{{{\rm{x}}^2} - {{\rm{a}}^2}}}$.

= $\frac{1}{{{{\rm{x}}^2} + {{\rm{a}}^2}}}\left[ {\frac{{{\rm{x}}\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} }} - \frac{{{\rm{x}}\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} }}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}} \right]$

= $\frac{1}{{{{\rm{x}}^2} - {{\rm{a}}^2}}}\left[ {\frac{{{\rm{x}}\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right) - {\rm{x}}\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}}{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} .\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}} \right]$

= $\frac{1}{{{{\rm{x}}^2} - {{\rm{a}}^2}}}\left[ {\frac{{{{\rm{x}}^3} - {{\rm{a}}^2}{\rm{x}} - {{\rm{x}}^3} - {{\rm{a}}^2}{\rm{x}}}}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} .\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} }}} \right]$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{2{{\rm{a}}^2}{\rm{x}}}}{{{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{\frac{3}{2}}}.\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} }}$.

 

6. Use the chain rule to calculate $\frac{{{\rm{dy}}}}{{{\rm{du}}}}$, if

(i)y = 2u² – 3u + 1 and u = 2x²

Solution:

Or, y = 2u² – 3u + 1

Differentiating both sides w.r.t. ‘u’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{du}}}}$ = $\frac{{\rm{d}}}{{{\rm{du}}}}$(2u² – 3u + 1)

= 2 * 2u – 3 * 1 + 0 = 4u – 3.

Again, u = 2x²

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(2x²) = 2 * 2x = 4x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{dy}}}}{{{\rm{du}}}}.\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ = (4u – 3).4x = 4x(4.2x² – 1) = 4x(8x² – 1).

 

(ii)y = 2u² + 3 and u = 3x² – 1

Solution:

Or, y = 2u² + 3

Differentiating both sides w.r.t. ‘u’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{du}}}}$ = $\frac{{\rm{d}}}{{{\rm{du}}}}$(2u² + 3) = 2 * 2u + 0 = 4u.

Again, u = 3x² – 1

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3x² – 1) = 3 * 2x – 0 = 6x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{dy}}}}{{{\rm{du}}}}.\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ = 4u.6x = 24x(3x² – 1).

 

(iii)y = 5t² + 6t – 7 and  t = x³ – 2

Solution:

Or, y = 5t² + 6t – 7

Differentiating both sides w.r.t. ‘t’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = $\frac{{\rm{d}}}{{{\rm{dt}}}}$(5t² + 6t – 7) = 10t + 6.

Again, t = x³ – 2

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x² – 2) = 3x².          

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{dy}}}}{{{\rm{du}}}}.\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ = (10t + 6).3x² = {10(x³ – 2) + 6}3x².

= 3x²(10x³ – 14) = 6x²(5x³ – 7).

 

(iv) y = $\frac{{{\rm{t}} - 1}}{{2{\rm{t}}}} = \frac{{\rm{t}}}{{2{\rm{t}}}} - \frac{1}{{2{\rm{t}}}} = \frac{1}{2} - \frac{1}{2}{{\rm{t}}^{ - 1}}$ and t = $\sqrt {{\rm{x}} + 1} $

Solution:

Or, y = $\frac{{{\rm{t}} - 1}}{{2{\rm{t}}}} = \frac{{\rm{t}}}{{2{\rm{t}}}} - \frac{1}{{2{\rm{t}}}} = \frac{1}{2} - \frac{1}{2}{{\rm{t}}^{ - 1}}$

Differentiating both sides w.r.t. ‘t’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = $\frac{{\rm{d}}}{{{\rm{dt}}}}$$\left( {\frac{1}{2} - \frac{1}{2}{{\rm{t}}^{ - 1}}} \right)$ =$ - \frac{1}{2}\left( { - 1} \right){{\rm{t}}^{ - 2}}$ = $\frac{1}{{2{{\rm{t}}^2}}}$.

Again, t = $\sqrt {{\rm{x}} + 1} $ = (x + 1)^1/2

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}{{\left( {{\rm{x}} + 1} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {{\rm{x}} + 1} \right)}}.\frac{{{\rm{d}}\left( {{\rm{x}} + 1} \right)}}{{{\rm{dx}}}}$ = $\frac{1}{2}{\left( {{\rm{x}} + 1} \right)^{ - \frac{1}{2}}}.1{\rm{\: }}$= $\frac{1}{{2\sqrt {{\rm{x}} + 1} }}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}.\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ = $\frac{1}{{2{{\rm{t}}^2}}}.\frac{1}{{2\sqrt {{\rm{x}} + 1} }}$ = $\frac{1}{{4{{\left( {\sqrt {{\rm{x}} + 1} } \right)}^2}.\sqrt {{\rm{x}} + 1} }} = \frac{1}{{4{{\left( {{\rm{x}} + 1} \right)}^{\frac{3}{2}}}}}$.

 

(v)  y = (2u² + 3)^1/3 and u = $\sqrt {\left( {2{\rm{x}} + 1} \right)} $

Solution:

Or, y = (2u² + 3)^1/3

Differentiating both sides w.r.t. ‘u’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{du}}}}$ = $\frac{{{\rm{d}}{{\left( {2{{\rm{u}}^2} + 3} \right)}^{\frac{1}{3}}}}}{{{\rm{du}}}}$ = $\frac{{{\rm{d}}{{\left( {2{{\rm{u}}^2} + 3} \right)}^{\frac{1}{3}}}}}{{{\rm{d}}\left( {2{{\rm{u}}^2} + 3} \right)}}.\frac{{{\rm{d}}\left( {2{{\rm{u}}^2} + 3} \right)}}{{{\rm{du}}}}$. = $\frac{1}{3}{\left( {2{{\rm{u}}^2} + 3} \right)^{ - \frac{2}{3}}}.$2.2u = $\frac{{4{\rm{u}}}}{{3{{\left( {2{{\rm{u}}^2} + 3} \right)}^{\frac{2}{3}}}}}$.

Again, u = $\sqrt {\left( {2{\rm{x}} + 1} \right)} $ = (2x + 1)^1/2

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}{{\left( {2{\rm{x}} + 1} \right)}^{\frac{1}{2}}}}}{{{\rm{dx}}}} = \frac{{{\rm{d}}{{\left( {2{\rm{x}} + 1} \right)}^{\frac{1}{2}}}}}{{{\rm{d}}\left( {2{\rm{x}} + 1} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{x}} + 1} \right)}}{{{\rm{dx}}}}$ = $\frac{1}{2}{\left( {2{\rm{x}} + 1} \right)^{ - \frac{1}{2}}}.1.2{\rm{\: }}$= $\frac{1}{{\sqrt {2{\rm{x}} + 1} }}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{dy}}}}{{{\rm{du}}}}.\frac{{{\rm{du}}}}{{{\rm{dx}}}}$ = $\frac{{4{\rm{u}}}}{{3{{\left( {2{{\rm{u}}^2} + 3} \right)}^{\frac{2}{3}}}}}.\frac{1}{{\sqrt {2{\rm{x}} + 1} }}$ = $\frac{{4\sqrt {2{\rm{x}} + 1} }}{{3{{\left\{ {2{{\left( {2{\rm{x}} + 1} \right)}^2} + 3} \right\}}^{\frac{2}{3}}}}}.\frac{1}{{\sqrt {2{\rm{x}} + 1} }} = \frac{4}{{3{{\left\{ {2\left( {2{\rm{x}} + 1} \right) + 3} \right\}}^{\frac{2}{3}}}}}$ = $\frac{4}{{3{{\left( {4{\rm{x}} + 5} \right)}^{\frac{2}{3}}}}}$.

 

(vi) y = $\frac{1}{{{{\rm{t}}^2} - 1}}$ and t = 3x² + 1

Solution:

Or, y = $\frac{1}{{{{\rm{t}}^2} - 1}}$

Differentiating both sides w.r.t. ‘t’.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = $\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {\frac{1}{{{{\rm{t}}^2} - 1}}} \right)$ = $\frac{{\left( {{{\rm{t}}^2} - 1} \right)\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{t}} - {\rm{t}}\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{{\rm{t}}^2} - 1} \right)}}{{{{\left( {{{\rm{t}}^2} - 1} \right)}^2}}}$ = $\frac{{\left( {{{\rm{t}}^2} - 1} \right).1 - {\rm{t}}.2{\rm{t}}}}{{{{\left( {{{\rm{t}}^2} - 1} \right)}^2}}}$ = $\frac{{{{\rm{t}}^2} - 1 - 2{{\rm{t}}^2}}}{{{{\left( {{{\rm{t}}^2} - 1} \right)}^2}}}$ = $\frac{{ - {{\rm{t}}^2} - 1}}{{{{\left( {{{\rm{t}}^2} - 1} \right)}^2}}}$.

Again, t = 3x² + 1

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {3{{\rm{x}}^2} + 1} \right)$ = 3 * 2x = 6x.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}.\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$ = $\frac{{ - {{\rm{t}}^2} - 1}}{{{{\left( {{{\rm{t}}^2} - 1} \right)}^2}}}$.6x = $\frac{{ - \left( {{{\rm{t}}^2} + 1} \right)}}{{{{\left( {{{\rm{t}}^2} - 1} \right)}^2}}}.6{\rm{x}}$ = $\frac{{\left\{ {{{\left( {3{{\rm{x}}^2} + 1} \right)}^2} + 1} \right\}.6{\rm{x}}}}{{{{\left\{ {{{\left( {3{{\rm{x}}^2} + 1} \right)}^2} - 1} \right\}}^2}}}$ = $\frac{{ - \left( {9{{\rm{x}}^4} + 6{{\rm{x}}^2} + 2} \right)6{\rm{x}}}}{{{{\left( {9{{\rm{x}}^4} + 6{{\rm{x}}^2}} \right)}^2}}}$.

= $\frac{{ - \left( {9{{\rm{x}}^4} + 6{{\rm{x}}^2} + 2} \right)6{\rm{x}}}}{{9{{\rm{x}}^4}{{\left( {3{{\rm{x}}^2} + 2} \right)}^2}}}$ = $\frac{{ - 2\left( {9{{\rm{x}}^4} + 6{{\rm{x}}^2} + 2} \right)}}{{3{{\rm{x}}^3}{{\left( {3{{\rm{x}}^2} + 2} \right)}^2}}}$.

7.Find the second derivative of the following:

(i)y = 7x² + 6x – 5

Solution:

y = 7x² + 6x – 5
Differentiating both sides w.r.t. ‘x’

or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(7x² + 6x – 5) = 14x + 6.

Again, differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(14x + 6)

Or, $\frac{{\left( {{{\rm{d}}^2}{\rm{y}}} \right)}}{{{\rm{d}}{{\rm{x}}^2}}}$ = 14.1 = 14.

 

(ii)y = 3x⁴ – x ² + 1

Solution:

y = 3x⁴ – x ² + 1
Differentiating both sides w.r.t. ‘x’

or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3x⁴ – x ² + 1) = 3.4x³ – 2x = 12x³ – 2x

Or, $\frac{{\left( {{{\rm{d}}^2}{\rm{y}}} \right)}}{{{\rm{d}}{{\rm{x}}^2}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(12x³ – 2x) = 36x² – 2.

 

(iii)y = $\frac{3}{{{{\rm{x}}^2}}}$

Solution:

y = $\frac{3}{{{{\rm{x}}^2}}}$
Differentiating both sides w.r.t. ‘x’

or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(3x^–2) = 3.(–2).x^–3.

Or, $\frac{{\left( {{{\rm{d}}^2}{\rm{y}}} \right)}}{{{\rm{d}}{{\rm{x}}^2}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(–6x^–3) = –6(–3).x^–4 = $\frac{{18}}{{{{\rm{x}}^4}}}$.

 

(iv)y = $\frac{1}{{2{\rm{x}} + 1}}$

Solution:

y = $\frac{1}{{2{\rm{x}} + 1}}$
Differentiating both sides w.r.t. ‘x’

or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(2x + 1)^–1 = $\frac{{{\rm{d}}{{\left( {2{\rm{x}} + 1} \right)}^{ - 1}}}}{{{\rm{d}}\left( {2{\rm{x}} + 1} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{x}} + 1} \right)}}{{{\rm{dx}}}}$ = –1.(2x + 1)^–2.2 = – 2(2x + 1)^–2.

Or, $\frac{{\left( {{{\rm{d}}^2}{\rm{y}}} \right)}}{{{\rm{d}}{{\rm{x}}^2}}}$ = $ - 2\frac{{\rm{d}}}{{{\rm{dx}}}}$(2x + 1)^–2 = –2$\frac{{{\rm{d}}{{\left( {2{\rm{x}} + 1} \right)}^{ - 2}}}}{{{\rm{d}}\left( {2{\rm{x}} + 1} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{x}} + 1} \right)}}{{{\rm{dx}}}}$ = –2.(–2).(2x + 1)^–3.2 = $\frac{8}{{{{\left( {2{\rm{x}} + 1} \right)}^3}}}$.

 

8. Use the implicit differentation to obtain $\frac{{{\rm{dy}}}}{{{\rm{du}}}}$ in the following:

(i)X² + y² = 16

Solution:

X² + y² = 16

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2} + {{\rm{y}}^2}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {16} \right)$

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right) + \frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0.

Or, 2x + 2y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0

Or, 2y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = –2x

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{2{\rm{x}}}}{{2{\rm{y}}}}$ = $ - \frac{{\rm{x}}}{{\rm{y}}}$.

 

(ii)$\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1

Solution:

Or, $\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}$ = 1

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} - \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( 1 \right)$

Or, $\frac{1}{{{{\rm{a}}^2}}}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right) - \frac{1}{{{{\rm{b}}^2}}}.\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0.

Or, $\frac{1}{{{{\rm{a}}^2}}}.2{\rm{x}} - \frac{1}{{{{\rm{b}}^2}}}.2{\rm{y}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}}{{\rm{b}}^2}}}{{2{\rm{y}}{{\rm{a}}^2}}}$ = $\frac{{{{\rm{b}}^2}{\rm{x}}}}{{{{\rm{a}}^2}{\rm{y}}}}$

(iii) y² = 4ax

Solution:

Or, y² = 4ax

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{y}}^2}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {4{\rm{ax}}} \right)$

Or, $\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = 4{\rm{a}}.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)$

Or, $2{\rm{y}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 4a.1

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{4{\rm{a}}}}{{2{\rm{y}}}}$ = $\frac{{2{\rm{a}}}}{{\rm{y}}}$

 

(iv)2x² + 3xy + 2y² = 0

Solution:

Or, 2x² + 3xy + 2y² = 0

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {2{{\rm{x}}^2}{\rm{\: }} + {\rm{\: }}3{\rm{xy\: }} + {\rm{\: }}2{{\rm{y}}^2}{\rm{\: }}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( 0 \right)$

Or, $2\frac{{\rm{d}}}{{{\rm{dx}}}}.\left( {{{\rm{x}}^2}} \right) + 3.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{xy}}} \right) + 2\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dx}}}}$ = 0

Or, 2.2x + 3 $\left\{ {{\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}}\frac{{{\rm{dx}}}}{{{\rm{dx}}}}} \right\}$ + 2 $\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0

Or, 4x + 3x $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 3y.1 + 2.2y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 0

Or, (3x + 4y) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = –4x – 3y

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\left( { - \frac{{4{\rm{x}} + 3{\rm{y}}}}{{3{\rm{x}} + 4{\rm{y}}}}} \right)$

 

(v)x² + 2x²y = y³

Solution:

Or, x² + 2x²y = y³

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2} + 2{{\rm{x}}^2}{\rm{y\: }}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{y}}^3}} \right)$

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}.\left( {{{\rm{x}}^2}} \right) + 2.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}{\rm{y}}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}$y³

Or, 2x + 2 $\left\{ {{{\rm{x}}^2}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right)} \right\}$ = $\frac{{{\rm{d}}{{\rm{y}}^3}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, 2x + 2x².$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 4xy = 3y²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.

Or, 2x + 4xy = (3y² – 2x²) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\left( {\frac{{2{\rm{x}}\left( {1 + 2{\rm{y}}} \right)}}{{3{{\rm{y}}^2} - 2{{\rm{x}}^2}}}} \right)$

 

(vi)x³y⁶ = (x + y)⁹

Solution:

Or, x³y⁶ = (x + y)⁹

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^3}{{\rm{y}}^6}{\rm{\: }}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}{\left( {{\rm{x}} + {\rm{y}}} \right)^9}$

Or, ${{\rm{x}}^3}\frac{{{\rm{d}}{{\rm{y}}^6}}}{{{\rm{dy}}}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{y}}^6}\frac{{{\rm{d}}{{\rm{x}}^3}}}{{{\rm{dx}}}} = \frac{{{\rm{d}}{{\left( {{\rm{x}} + {\rm{y}}} \right)}^9}}}{{{\rm{d}}\left( {{\rm{x}} + {\rm{y}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{x}} + {\rm{y}}} \right)}}{{{\rm{dx}}}}$.

Or, x³.6y⁵$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y⁶.3x² = 9(x + y)⁸$\left\{ {1 + \frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right\}$.

Or, 6x³y⁵$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 3x²y⁶ = 9(x + y)⁸ + 9(x + y)⁸$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.

Or, {6x³y⁵$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 3x²y⁶ = 9(x + y)⁸ + 9(x + y)⁸$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$}

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{9{{\left( {{\rm{x}} + {\rm{y}}} \right)}^8} - 3{{\rm{x}}^2}{{\rm{y}}^6}}}{{6{{\rm{x}}^3}{{\rm{y}}^5} - 9{{\left( {{\rm{x}} + {\rm{y}}} \right)}^8}}}$.

= $\frac{{\frac{{9{{\left( {{\rm{x}} + {\rm{y}}} \right)}^9}}}{{{\rm{x}} + {\rm{y}}}} - 3{{\rm{x}}^2}{{\rm{y}}^6}}}{{6{{\rm{x}}^3}{{\rm{y}}^5} - \frac{{9{{\left( {{\rm{x}} + {\rm{y}}} \right)}^9}}}{{{\rm{x}} + {\rm{y}}}}}}$ = $\frac{{9\frac{{{{\rm{x}}^3}{{\rm{y}}^6}}}{{{\rm{x}} + {\rm{y}}}} - 3{{\rm{x}}^2}{{\rm{y}}^6}}}{{6{{\rm{x}}^3}{{\rm{y}}^5} - 9\frac{{{{\rm{x}}^3}{{\rm{y}}^6}}}{{{\rm{x}} + {\rm{y}}}}}}$$\left[ {{{\rm{x}}^3}{{\rm{y}}^6} = {{\left( {{\rm{x}} + {\rm{y}}} \right)}^9}} \right]$.

= $\frac{{9{{\rm{x}}^3}{{\rm{y}}^6} - 3{{\rm{x}}^3}{{\rm{y}}^6} - 3{{\rm{x}}^2}{{\rm{y}}^7}}}{{6{{\rm{x}}^4}{{\rm{y}}^5} + 6{{\rm{x}}^3}{{\rm{y}}^6} - 9{{\rm{x}}^3}{{\rm{y}}^6}}}$

= $\frac{{6{{\rm{x}}^3}{{\rm{y}}^6} - 3{{\rm{x}}^3}{{\rm{y}}^7}}}{{6{{\rm{x}}^4}{{\rm{y}}^5} - 3{{\rm{x}}^3}{{\rm{y}}^6}}}$ = $\frac{{3{{\rm{x}}^2}{{\rm{y}}^6}\left( {2{\rm{x}} - {\rm{y}}} \right)}}{{3{{\rm{x}}^3}{{\rm{y}}^5}\left( {2{\rm{x}} - {\rm{y}}} \right)}}$ = $\frac{{\rm{y}}}{{\rm{x}}}$.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\left( {\frac{{2{\rm{x}}\left( {1 + 2{\rm{y}}} \right)}}{{3{{\rm{y}}^2} - 2{{\rm{x}}^2}}}} \right)$

 

(vii) x²y+ xy ²= a³

Solution:

Or, x²y+ xy ²= a³

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}{\rm{y\: }} + {\rm{\: x}}{{\rm{y}}^2}{\rm{\: }}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}{\left( {\rm{a}} \right)^3}$

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}{\rm{y}}} \right) + \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}}{{\rm{y}}^2}} \right)$ = 0.

Or, x².$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x²) + x$\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}$. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y²$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x) = 0

Or, x²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y.2x + x.2y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y².1 = 0

Or,  x(x + 2y) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = –2xy – y².

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{{\rm{y}}\left( {2{\rm{x}} + {\rm{y}}} \right)}}{{{\rm{x}}\left( {{\rm{x}} + 2{\rm{y}}} \right)}}$

 

(viii)x³ + y³ = 3xy²

Solution:

Or, x³ + y³ = 3xy²

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^3} + {{\rm{y}}^3}{\rm{\: }}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {3{\rm{x}}{{\rm{y}}^2}} \right)$

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^3}} \right) + \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{y}}^3}} \right)$ = 3. $\frac{{\rm{d}}}{{{\rm{dx}}}}$(xy²)

Or, 3x² + $\frac{{{\rm{d}}{{\rm{y}}^3}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 3$\left\{ {{\rm{x}}\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{y}}^2}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)} \right\}$.

Or, 3x² + 3y²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 6xy $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 3y²

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{3{{\rm{y}}^2} - 3{{\rm{x}}^2}}}{{3{{\rm{y}}^2} - 6{\rm{xy}}}}$ = $\frac{{3\left( {{{\rm{y}}^2} - {{\rm{x}}^2}} \right)}}{{3{\rm{y}}\left( {{\rm{y}} - 2{\rm{x}}} \right)}}$ = $\frac{{{{\rm{y}}^2} - {{\rm{x}}^2}}}{{{\rm{y}}\left( {{\rm{y}} - 2{\rm{x}}} \right)}}$

 

(ix) x²y² = x² + y²

Solution:

Or, x²y² = x² + y²

Differentiating both sides w.r.t. ‘x’.

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}{{\rm{y}}^2}{\rm{\: }}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2} + {{\rm{y}}^2}} \right)$

Or, ${{\rm{x}}^2}\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}.}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{y}}^2}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right)$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right) + \frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, x².2y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y².2x = 2x + 2y. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, 2y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$(x² – 1) = 2x(1 – y²)

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ =$\frac{{{\rm{x}}\left( {1 - {{\rm{y}}^2}} \right)}}{{{\rm{y}}\left( {{{\rm{x}}^2} - 1} \right)}}$.

 

9.(i) Differentiate x⁶ w.r.t. to x².

Solution:

Let y = x⁶

Diff. both sides w.r.t. to x².

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}{{\rm{x}}^6}}}{{{\rm{d}}{{\rm{x}}^2}}}$ = $\frac{{\frac{{{\rm{d}}{{\rm{x}}^6}}}{{{\rm{dx}}}}}}{{\frac{{{\rm{d}}{{\rm{x}}^2}}}{{{\rm{dx}}}}}}$ = $\frac{{6{{\rm{x}}^5}}}{{2{\rm{x}}}}$ = 3x⁴.

 

(ii)Differentiate (3x + 1)⁴ w.r.t. to x²

Solution:

Let y = (3x + 1)⁴

Diff. both sides w.r.t. to x²

Or, $\frac{{{\rm{dy}}}}{{{\rm{d}}\left( {3{\rm{x}} + 1} \right)}}$ = $\frac{{{\rm{d}}{{\left( {3{\rm{x}} + 1} \right)}^4}}}{{{\rm{d}}\left( {3{\rm{x}} + 1} \right)}}$ = 4(3x + 1)³.

 

(iii)Differentiate 2x⁶ – 3x⁴ + x²  w.r.t. x³.

Solution:

Let y = 2x⁶ – 3x⁴ + x²

Diff. both sides w.r.t. x³

Or, $\frac{{{\rm{dy}}}}{{{\rm{d}}{{\rm{x}}^3}}}$ = $\frac{{{\rm{d}}\left( {2{{\rm{x}}^6} - 3{{\rm{x}}^4} + {{\rm{x}}^2}} \right)}}{{{\rm{d}}{{\rm{x}}^3}}}$

= $\frac{{{\rm{d}}\left( {2{{\rm{x}}^6}} \right)}}{{{\rm{d}}{{\rm{x}}^3}}} - \frac{{{\rm{d}}\left( {3{{\rm{x}}^4}} \right)}}{{{\rm{d}}{{\rm{x}}^3}}} + \frac{{{\rm{d}}{{\rm{x}}^2}}}{{{\rm{d}}{{\rm{x}}^3}}}$ = $\frac{{2.\frac{{{\rm{d}}{{\rm{x}}^6}}}{{{\rm{dx}}}}}}{{\frac{{{\rm{d}}{{\rm{x}}^3}}}{{{\rm{dx}}}}}} - 3.\frac{{\frac{{{\rm{d}}{{\rm{x}}^4}}}{{{\rm{dx}}}}}}{{\frac{{{\rm{d}}{{\rm{x}}^3}}}{{{\rm{dx}}}}}} + \frac{{\frac{{{\rm{d}}{{\rm{x}}^2}}}{{{\rm{dx}}}}}}{{\frac{{{\rm{d}}{{\rm{x}}^3}}}{{{\rm{dx}}}}}}$

= $\frac{{2.6{{\rm{x}}^5}}}{{3{{\rm{x}}^2}}} - \frac{{3.4}}{{3{{\rm{x}}^3}}} + \frac{{2{\rm{x}}}}{{3{{\rm{x}}^2}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 4x³ – 4x + $\frac{2}{{3{\rm{x}}}}$.

 

(iv) Differentiate (5x – 1)⁶ w.r.t. (x + 1)².

Solution:

Let y = (5x – 1)⁶

Diff. both sides w.r.t. (x + 1)²

Or, $\frac{{{\rm{dy}}}}{{{\rm{d}}{{\left( {{\rm{x}} + 1} \right)}^2}}}$ = $\frac{{{\rm{d}}{{\left( {5{\rm{x}} - 1} \right)}^6}}}{{{\rm{d}}{{\left( {{\rm{x}} + 1} \right)}^2}}}$

= $\frac{{\frac{{{\rm{d}}{{\left( {5{\rm{x}} - 1} \right)}^6}}}{{{\rm{d}}\left( {5{\rm{x}} - 1} \right)}}.\frac{{{\rm{d}}\left( {5{\rm{x}} - 1} \right)}}{{{\rm{dx}}}}}}{{\frac{{{\rm{d}}{{\left( {{\rm{x}} + 1} \right)}^2}}}{{{\rm{d}}\left( {{\rm{x}} + 1} \right)}}.\frac{{{\rm{d}}\left( {{\rm{x}} + 1} \right)}}{{{\rm{dx}}}}}}$

= $\frac{{6{{\left( {5{\rm{x}} - 1} \right)}^5}.5}}{{2\left( {{\rm{x}} + 1} \right).1}}$ = $\frac{{15{{\left( {5{\rm{x}} - 1} \right)}^5}}}{{{\rm{x}} + 1}}$.

So, $\frac{{{\rm{dy}}}}{{{\rm{d}}{{\left( {{\rm{x}} + 1} \right)}^2}}}$ = $\frac{{15{{\left( {5{\rm{x}} - 1} \right)}^5}}}{{{\rm{x}} + 1}}$.