Derivatives Exercise: 17.3 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 17.3

1.Find, From the first principle the derivatives of:

(i)log(ax + b)

Solution:

Let y = log(ax + b)

Let $\Delta {\rm{x}}$ and $\Delta {\rm{y}}$ be the small increments in x and y respectively. Then,

Or, y + $\Delta {\rm{y}}$ = log(a(x + $\Delta {\rm{x}}$) + b)

Or, $\Delta {\rm{y}}$ = log(ax + b + a$\Delta {\rm{x}}$) – log(ax + b)

= log $\frac{{{\rm{ax}} + {\rm{b}} + {\rm{a}}\Delta {\rm{x}}}}{{{\rm{ax}} + {\rm{b}}}}$ = log $\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{{\rm{ax}} + {\rm{b}}}} + \frac{{{\rm{a}}\Delta {\rm{x}}}}{{{\rm{ax}} + {\rm{b}}}}} \right)$ = log $\left( {1 + \frac{{{\rm{a}}\Delta {\rm{x}}}}{{{\rm{ax}} + {\rm{b}}}}} \right)$.

Or, $\Delta {\rm{y}}$ = $\frac{{{\rm{a}}\Delta {\rm{x}}}}{{{\rm{ax}} + {\rm{b}}}} - \frac{{{{\left( {\frac{{{\rm{a}}\Delta {\rm{x}}}}{{{\rm{ax}} + {\rm{b}}}}} \right)}^2}}}{2}$ + …. Higher power of $\Delta {\rm{x}}$.

= $\frac{{{\rm{a}}\Delta {\rm{x}}}}{{{\rm{ax}} + {\rm{b}}}}\left[ {1 - \frac{{{\rm{a}}\Delta {\rm{x}}}}{{2\left( {{\rm{x}} + {\rm{b}}} \right)}} +  \ldots .} \right]{\rm{\: }}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{\rm{a}}}{{{\rm{ax}} + {\rm{b}}}}$$\left[ {1 - \frac{{{\rm{a}}\Delta {\rm{x}}}}{{2\left( {{\rm{ax}} + {\rm{b}}} \right)}} +  \ldots } \right]$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{a}}}{{{\rm{ax}} + {\rm{b}}}}.1{\rm{\: }}$= $\frac{{\rm{a}}}{{{\rm{ax}} + {\rm{b}}}}$.

 

(ii)log₅ x

Solution:

Let y = log₅ x = log₅ e log(x + $\Delta {\rm{x}}$)

Let $\Delta {\rm{x}}$ and $\Delta {\rm{y}}$ be the small increments in x and y respectively. Then,

Or, y + $\Delta {\rm{y}}$ = log₅ e log(x + $\Delta {\rm{x}}$)

Or, $\Delta {\rm{y}}$ = log₅ e. log(x + $\Delta {\rm{x}}$) – log₅e.logx

= log₅e {log(x + $\Delta {\rm{x}}$) – logx}

= log₅e {log(x + $\Delta {\rm{x}}$) – logx}

= log₅e log $\frac{{{\rm{x}} + \Delta {\rm{x}}}}{{\rm{x}}}$ = log₅ e. log$\left( {1 + \frac{{\Delta {\rm{x}}}}{{\rm{x}}}} \right)$

= log₅e $\{ \frac{{\Delta {\rm{x}}}}{{\rm{x}}} - \frac{{{{\left( {\frac{{\Delta {\rm{x}}}}{{\rm{x}}}} \right)}^2}}}{2} + $..... higher power of $\Delta {\rm{x}}$}

Or, $\Delta {\rm{y}}$ = log₅e $\frac{{\Delta {\rm{x}}}}{{\rm{x}}}\left\{ {1 - \frac{{\Delta {\rm{x}}}}{{2{\rm{x}}}} +  \ldots .} \right\}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = log₅e $\frac{1}{{\rm{x}}}$$\left\{ {1 - \frac{{\Delta {\rm{x}}}}{{2{\rm{x}}}} +  \ldots } \right\}$

Or, $\Delta {\rm{x}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0$$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\Delta {\rm{x}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0$ log₅ e $\frac{1}{{\rm{x}}}$$\left[ {1 - \frac{{\Delta {\rm{x}}}}{{2{\rm{x}}}} +  \ldots .} \right]$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = log₅e $\frac{1}{{\rm{x}}}$.1 = log₅e $\frac{1}{{\rm{x}}}$.

 

(iii)log $\frac{{\rm{x}}}{{10}}$

Solution:

Let y = log $\frac{{\rm{x}}}{{10}}$

Let $\Delta {\rm{x}}$ and $\Delta {\rm{y}}$ be the small increments in x and y respectively. Then,

Or, y + $\Delta {\rm{y}}$ = log$\frac{{{\rm{x}} + \Delta {\rm{x}}}}{{10}}$

Or, $\Delta {\rm{y}}$ = log $\frac{{{\rm{x}} + \Delta {\rm{x}}}}{{10}} - \log \frac{{\rm{x}}}{{10}}$ = log $\frac{{\frac{{{\rm{x}} + \Delta {\rm{x}}}}{{10}}}}{{\frac{{\rm{x}}}{{10}}}}$

= log $\left( {1 + \frac{{\Delta {\rm{x}}}}{{\rm{x}}}} \right)$ = $\frac{{\Delta {\rm{x}}}}{{\rm{x}}} - \frac{{{{\left( {\frac{{\Delta {\rm{x}}}}{{\rm{x}}}} \right)}^2}}}{2} + $….. higher power of $\Delta {\rm{x}}$.

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

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{1}{{\rm{x}}}\left\{ {1 - \frac{{\Delta {\rm{x}}}}{{2{\rm{x}}}} +  \ldots .} \right\}$

Or, $\Delta {\rm{x}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0$$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\Delta {\rm{x}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0$ log₅ e $\frac{1}{{\rm{x}}}$$\left[ {1 + \frac{{\Delta {\rm{x}}}}{{2{\rm{x}}}} +  \ldots .} \right]$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{1}{{\rm{x}}}$.1 = $\frac{1}{{\rm{x}}}$.

 

(iv)e^ax+b

Solution:

Let y = e^ax+b

Let $\Delta {\rm{x}}$ and $\Delta {\rm{y}}$ be the small increments in x and y respectively. Then,

Or, y + $\Delta {\rm{y}}$ = ${{\rm{e}}^{{\rm{a}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right) + {\rm{b}}}}$

Or, $\Delta {\rm{y}}$ = ${{\rm{e}}^{{\rm{ax}} + {\rm{b}} + {\rm{a}}\Delta {\rm{x}}}} - {{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}$

= e^{ax + b}.${{\rm{e}}^{{\rm{a}}\Delta {\rm{x}}}}$ – e^ax+b = e^{ax + b}(${{\rm{e}}^{{\rm{a}}\Delta {\rm{x}}}}$ – 1)

= e^{ax +b} $\left\{ {1 + \frac{{{\rm{a}}\Delta {\rm{x}}}}{{1!}} + \frac{{{{\left( {{\rm{a}}\Delta {\rm{x}}} \right)}^2}}}{{2!}} +  \ldots . - 1} \right\}$

= e^{ax + b}$\left\{ {{\rm{a}}\Delta {\rm{x}} + \frac{{{{\left( {{\rm{a}}\Delta {\rm{x}}} \right)}^2}}}{{2!}} +  \ldots } \right\}$

Or, $\Delta {\rm{y}}$ = e^{ax + b}. a$\Delta {\rm{x}}$$\left\{ {1 + \frac{{{\rm{a}}\Delta {\rm{x}}}}{{2!}} +  \ldots } \right\}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = e^{ax + b}.a$\left\{ {1 + \frac{{{\rm{a}}\Delta {\rm{x}}}}{{2!}} +  \ldots .} \right\}$

Or, $\Delta {\rm{x}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0$$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\Delta {\rm{x}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0$ e^{ax + b} $\left\{ {1 + \frac{{{\rm{a}}\Delta {\rm{x}}}}{{2!}} +  \ldots } \right\}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = e^{ax + b}.a . 1 = a.e^{ax + b}.

 

(v)e^x/3

Solution:

Let y = e^x/3

Let $\Delta {\rm{x}}$ and $\Delta {\rm{y}}$ be the small increments in x and y respectively. Then,

Or, y + $\Delta {\rm{y}}$ = ${{\rm{e}}^{\frac{{{\rm{x}} + \Delta {\rm{x}}}}{3}}}$

Or, $\Delta {\rm{y}}$ = ${{\rm{e}}^{\frac{{{\rm{x}} + \Delta {\rm{x}}}}{3}}} - {{\rm{e}}^{\frac{{\rm{x}}}{3}}}$ = ${{\rm{e}}^{\frac{{\rm{x}}}{3} + \frac{{\Delta {\rm{x}}}}{3}}} - {{\rm{e}}^{\frac{{\rm{x}}}{3}}}$

= e^x/3.${{\rm{e}}^{\frac{{\Delta {\rm{x}}}}{3}}}$ – e^x/3 = e^x/3(${{\rm{e}}^{\frac{{\Delta {\rm{x}}}}{3}}}$ – 1)

= e^x/3 $\left\{ {1 + \frac{{\Delta {\rm{x}}}}{3} + \frac{{{{\left( {\Delta {\rm{x}}/3} \right)}^2}}}{{2!}} +  \ldots . - 1} \right\}$

= e^x/3$\left\{ {\frac{{\Delta {\rm{x}}}}{3} + \frac{{{{\left( {\frac{{\Delta {\rm{x}}}}{3}} \right)}^2}}}{{2!}} +  \ldots } \right\}$

Or, $\Delta {\rm{y}}$ = e^x/3. $\frac{{\Delta {\rm{x}}}}{3}$$\left\{ {1 + \frac{{\frac{{\Delta {\rm{x}}}}{3}}}{{2!}} +  \ldots } \right\}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{1}{3}$ e^1/3$\left\{ {1 + \frac{{\frac{{\Delta {\rm{x}}}}{3}}}{{2!}} +  \ldots .} \right\}$

Or, $\Delta {\rm{x}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0$$\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\Delta {\rm{x}}\begin{array}{*{20}{c}}{{\rm{lim}}}\\\to\end{array}0$.e^x/3$\left\{ {1 + \frac{{\frac{{\Delta {\rm{x}}}}{3}}}{{2!}} +  \ldots .} \right\}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{3}$.e^x/3.1 = $\frac{1}{3}$e^x/3.

 

2. Find the derivative of:

(i)log(sinx)

Solution:

Let y = log(sinx)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(log.sinx) = $\frac{{{\rm{d}}\left( {{\rm{logsinx}}} \right)}}{{{\rm{d}}\left( {{\rm{sinx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sinx}}} \right)}}{{{\rm{dx}}}}$

= $\frac{1}{{\sin {\rm{x}}}}$.cosx = cotx.

 

(ii)log(x + tanx)

Solution:

Let y = log(x + tanx)

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

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

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

 

(iii)log(1 + e^5x)

Solution:

Let y = log(1 + e^5x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(log(1 + e^5x)) = $\frac{{{\rm{d}}\left( {{\rm{log}}\left( {1 + {{\rm{e}}^{5{\rm{x}}}}} \right)} \right)}}{{{\rm{d}}\left( {1 + {{\rm{e}}^{5{\rm{x}}}}} \right)}}.\frac{{{\rm{d}}\left( {1 + {{\rm{e}}^{5{\rm{x}}}}} \right)}}{{{\rm{dx}}}}$

= $\frac{1}{{1 + {{\rm{e}}^{5{\rm{x}}}}}}\left\{ {\frac{{\rm{d}}}{{{\rm{dx}}}}\left( 1 \right) + \frac{{{\rm{d}}\left( {{{\rm{e}}^{5{\rm{x}}}}} \right)}}{{{\rm{d}}\left( {5{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {5{\rm{x}}} \right)}}{{{\rm{dx}}}}} \right\}$ = $\frac{1}{{1 + {{\rm{e}}^{5{\rm{x}}}}}}\left( {0 + {{\rm{e}}^{5{\rm{x}}}}.5.1} \right)$ = $\frac{{5{{\rm{e}}^{5{\rm{x}}}}}}{{1 + {{\rm{e}}^{5{\rm{x}}}}}}$.

 

(iv) log(log x)

Solution:

Let y = log(log x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(log(logx)) = $\frac{{{\rm{d}}\left( {{\rm{log}}\left( {{\rm{logx}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{logx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{logx}}} \right)}}{{{\rm{dx}}}}$

= $\frac{1}{{{\rm{logx}}}}.\frac{1}{{\rm{x}}} = \frac{1}{{{\rm{xlogx}}}}$.

 

(v) log(sec x)

Solution:

Let y = log(sec x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(log(sec x)) = $\frac{{{\rm{d}}\left( {{\rm{log}}\left( {\sec {\rm{x}}} \right)} \right)}}{{{\rm{d}}\left( {\sec {\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{secx}}} \right)}}{{{\rm{dx}}}}$

= $\frac{1}{{{\rm{secx}}}}.$sec x.tan x = tanx.

 

(vi)log(1 + sin²x)

Solution:

Let y = log(1 + sin²x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(log(1 + sin²x)) = $\frac{{{\rm{d}}\left( {{\rm{log}}\left( {1 + {{\sin }^2}{\rm{x}}} \right)} \right)}}{{{\rm{d}}\left( {1 + {\rm{si}}{{\rm{n}}^2}{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {1 + {{\sin }^2}{\rm{x}}} \right)}}{{{\rm{dx}}}}$

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

= $\frac{1}{{1 + {{\sin }^2}{\rm{x}}}}$(0 + 2sinx.cosx) = $\frac{{2{\rm{sinx}}.{\rm{cosx}}}}{{1 + {{\sin }^2}{\rm{x}}}}$.

 

(vii)log(e^ax + e^–ax)

Solution:

Let y = log(e^ax + e^–ax)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(log(e^ax + e^–ax))= $\frac{{{\rm{d}}\left( {{\rm{log}}\left( {{{\rm{e}}^{{\rm{ax}}}}{\rm{\: }} + {\rm{\: }}{{\rm{e}}^{ - {\rm{ax}}}}} \right)} \right)}}{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{ax}}}} + {{\rm{e}}^{ - {\rm{ax}}}}} \right)}}.\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{ax}}}} + {{\rm{e}}^{ - {\rm{ax}}}}} \right)}}{{{\rm{dx}}}}$

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

= $\frac{1}{{{{\rm{e}}^{{\rm{ax}}}} + {{\rm{e}}^{ - {\rm{ax}}}}}}$(e^ax.a.1 + e^–ax.(–a).1) = $\frac{{{\rm{a}}\left( {{{\rm{e}}^{{\rm{ax}}}} - {{\rm{e}}^{ - {\rm{ax}}}}} \right)}}{{{{\rm{e}}^{{\rm{ax}}}} + {{\rm{e}}^{ - {\rm{ax}}}}}}$.

 

(viii)log$\left( {\sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}}  + {\rm{b}}} \right)$

Solution:

Let y = log$\left( {\sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}}  + {\rm{b}}} \right)$

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

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

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

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

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

 

(ix)log$\left( {\sqrt {{\rm{a}} + {\rm{x}}}  + \sqrt {{\rm{a}} - {\rm{x}}} } \right)$

Solution:

Let y = log$\left( {\sqrt {{\rm{a}} + {\rm{x}}}  + \sqrt {{\rm{a}} - {\rm{x}}} } \right)$

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

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

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

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

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

 

(x) ln|x – 4|

Solution:

Let y = ln|x – 4|

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left[ {\ln \left| {{\rm{x}} - 4} \right|} \right]}}{{{\rm{dx}}}}$    [if x – 4 > 0]

= $\frac{{{\rm{d}}.\ln \left( {{\rm{x}} - 4} \right)}}{{{\rm{d}}\left( {{\rm{x}} - 4} \right)}}.\frac{{{\rm{d}}\left( {{\rm{x}} - 4} \right)}}{{{\rm{dx}}}}$ = $\frac{1}{{{\rm{x}} - 4}}.1$ = $\frac{1}{{{\rm{x}} - 4}}$.

Again, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left[ {\ln \left| {{\rm{x}} - 4} \right|} \right]}}{{{\rm{dx}}}}$.

= $\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{ln}}\left( {4 - {\rm{x}}} \right)$    [If x – 4 < 0, then 4 – x > 0]

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

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{{\rm{x}} - 4}}$.

 

3. FInd the differential coefficient of:

(i)e^sinx

Solution:

Let y = e^sinx

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(e^sinx) = $\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{sinx}}}}} \right)}}{{{\rm{d}}\left( {{\rm{sinx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sinx}}} \right)}}{{{\rm{dx}}}}$ = e^sinx.cosx.

 

(ii)${{\rm{e}}^{\sqrt {{\rm{cosx}}} }}$

Solution:

Let y = ${{\rm{e}}^{\sqrt {{\rm{cosx}}} }}$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(${{\rm{e}}^{\sqrt {{\rm{cosx}}} }}$) = $\frac{{{\rm{d}}\left( {{{\rm{e}}^{\sqrt {{\rm{cosx}}} }}} \right)}}{{{\rm{d}}\left( {\sqrt {{\rm{cosx\: }}} } \right)}}.\frac{{{\rm{d}}\left( {\sqrt {{\rm{cosx}}} } \right)}}{{{\rm{d}}\left( {{\rm{cosx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cosx}}} \right)}}{{{\rm{dx}}}}$

= ${{\rm{e}}^{\sqrt {{\rm{cosx}}} }}.\frac{1}{{2\sqrt {{\rm{cosx}}} }}$.(–sinx) = $ - \frac{{{{\rm{e}}^{\sqrt {{\rm{cosx}}} }}.{\rm{sinx}}}}{{2\sqrt {{\rm{cosx}}} }}$.

 

(iii)${{\rm{e}}^{\left( {1 + {\rm{logx}}} \right)}}$

Solution:

Let y = ${{\rm{e}}^{\left( {1 + {\rm{logx}}} \right)}}$

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

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

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

 

(iv)${{\rm{e}}^{{\rm{sin}}\left( {{\rm{logx}}} \right)}}$

Solution:

Let y = ${{\rm{e}}^{{\rm{sin}}\left( {{\rm{logx}}} \right)}}$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(${{\rm{e}}^{{\rm{sin}}\left( {{\rm{logx}}} \right)}}$) = $\frac{{{\rm{d}}\left( {{{\rm{e}}^{\sin \left( {{\rm{logx}}} \right)}}} \right)}}{{{\rm{d}}\left( {\sin \left( {{\rm{logx}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {\sin \left( {{\rm{logx}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{logx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{logx}}} \right)}}{{{\rm{dx}}}}$

= ${{\rm{e}}^{{\rm{sin}}.\left( {{\rm{logx}}} \right)}}$.cos(logx).$\frac{1}{{\rm{x}}}$ = $\frac{1}{{\rm{x}}}$.e^sin(logx).cos(logx).

 

(v)tan(log x)

Solution:

Let y = tan(log x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan(log x))= $\frac{{{\rm{d}}\left( {{\rm{tan}}\left( {{\rm{log\: x}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{logx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{logx}}} \right)}}{{{\rm{dx}}}}$

= sec²(logx).$\frac{1}{{\rm{x}}}$ = $\frac{1}{{\rm{x}}}$sec²(logx).

 

(vi)sin(1 + e^ax)

Solution:

Let y = sin(1 + e^ax)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sin(1 + e^ax))= $\frac{{{\rm{d}}\left( {{\rm{sin}}\left( {1 + {{\rm{e}}^{{\rm{ax}}}}} \right)} \right)}}{{{\rm{d}}\left( {1 + {{\rm{e}}^{{\rm{ax}}}}} \right)}}.\frac{{{\rm{d}}\left( {1 + {{\rm{e}}^{{\rm{ax}}}}} \right)}}{{{\rm{dx}}}}$

= cos(1 + e^ax) $\left\{ {\frac{{\rm{d}}}{{{\rm{dx}}}}\left( 1 \right) + \frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}}} \right)}}{{{\rm{dx}}}}} \right\}$

= cos(1 + e^ax){0 + e^ax.a.1} = ae^ax.cos(1 + e^ax).

 

(vii)cos(log.secx)

Solution:

Let y = cos(log.secx)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(cos(log.secx)) = $\frac{{{\rm{d}}\left( {{\rm{cos}}\left( {{\rm{log}}.{\rm{secx}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{log}}.{\rm{secx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{log}}.{\rm{secx}}} \right)}}{{{\rm{d}}\left( {{\rm{secx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{secx}}} \right)}}{{{\rm{dx}}}}$

= – sin(log.secx).$\frac{1}{{{\rm{secx}}}}$.secx.tanx = – tanx.sin(log.sec x).

 

(viii) sec(log tan x)

Solution:

Let y = sec(log.tanx)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sec(log.tanx)) = $\frac{{{\rm{d}}\left( {{\rm{sec}}\left( {{\rm{log}}.{\rm{tanx}}} \right)} \right){\rm{\: }}}}{{{\rm{d}}\left( {{\rm{log}}.{\rm{tanx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{log}}.{\rm{tanx}}} \right)}}{{{\rm{d}}\left( {{\rm{tan}}.{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{tan}}.{\rm{x}}} \right)}}{{{\rm{dx}}}}$

= sec.(log.tanx).tan(log.tanx).$\frac{1}{{{\rm{tanx}}}}$.sec²x

= $\frac{{\sec \left( {{\rm{log}}.{\rm{tanx}}} \right).\tan \left( {{\rm{log}}.{\rm{tanx}}} \right){{\sec }^2}{\rm{x}}}}{{{\rm{tanx}}}}$.

 

(ix)sin.log.sin$.{{\rm{e}}^{{{\left( {\rm{x}} \right)}^2}}}$

Solution:

Let y = sin.log.sin$.{{\rm{e}}^{{{\left( {\rm{x}} \right)}^2}}}$

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

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

= cos.log.sin.${{\rm{e}}^{{{\rm{x}}^2}}}$. $\frac{1}{{\sin {{\rm{e}}^{{{\rm{x}}^2}}}}}$.cos ${{\rm{e}}^{{{\rm{x}}^2}}}$.${{\rm{e}}^{{{\rm{x}}^2}}}$.2x

= 2x.${{\rm{e}}^{{{\rm{x}}^2}}}$.cot ${{\rm{e}}^{{{\rm{x}}^2}}}$.cos.log.sin${{\rm{e}}^{{{\rm{x}}^2}}}{\rm{\: }}$.

 

4. Differentiate the following with respect to x:

(i)x²log(1 + x)

Solution:

Let y = x²log(1 + x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}${x².log(1 + x)}

= x².$\frac{{\rm{d}}}{{{\rm{dx}}}}$(log(1 + x)) + log(1 + x).$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x²)

= x². $\frac{{{\rm{d}}\left( {\log \left( {1 + {\rm{x}}} \right)} \right)}}{{{\rm{d}}\left( {1 + {\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {1 + {\rm{x}}} \right)}}{{{\rm{dx}}}} + \log \left( {1 + {\rm{x}}} \right).2{\rm{x}}$

= x².$\frac{1}{{1 + {\rm{x}}}}$.1 + 2xlog(1 + x) = $\frac{{{{\rm{x}}^2}}}{{1 + {\rm{x}}}}$ + 2x.log(1 + x).

 

(ii)x⁵.e^ax

Solution:

Let y = x⁵.e^ax

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x⁵.e^ax)

= x⁵.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x⁵.e^ax) = x⁵.$\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}}} \right)}}{{{\rm{dx}}}}$ + e^ax$\frac{{\rm{d}}}{{{\rm{dx}}}}$(x⁵)

= x⁵.e^ax.a.1 + e^ax.5x⁴ = 5x⁴.e^ax + ax⁵.e^ax.

 

(iii)sinax.logx

Solution:

Let y = sinax.logx

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sinax.logx)

= sinax.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(logx) = logx.$\frac{{{\rm{d}}\left( {{\rm{sinax}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}}} \right)}}{{{\rm{dx}}}}$

= $\frac{{{\rm{sinax}}}}{{\rm{x}}}$ + logx.cosax.a = $\frac{{{\rm{sinax}}}}{{\rm{x}}}$ + a.cosax.logx

 

(iv)e^ax.cosbx

Solution:

Let y = e^ax.cosbx

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(e^ax.cosbx)

= e^ax.$\frac{{{\rm{d}}\left( {{\rm{cosbx}}} \right)}}{{{\rm{d}}\left( {{\rm{bx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{bx}}} \right)}}{{{\rm{dx}}}}$ + cosbx.$\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}}} \right)}}{{{\rm{dx}}}}$

 = e^ax.(–sinbx).b.1 + cosbx.e^ax.a.1 = (a.cosbx – b.sinbx).e^ax.

 

(v)(tanx + x²).logx

Solution:

Let y = (tanx + x²).logx

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}${(tanx + x²).logx}

= (tanx + x²)$\frac{{\rm{d}}}{{{\rm{dx}}}}$(log x) + logx $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tanx + x²).

= (tanx + x²).$\frac{1}{{\rm{x}}}$ + logx.(sec²x + 2x)

= $\frac{{{\rm{tanx}} + {{\rm{x}}^2}}}{{\rm{x}}}$ + (Sec²x + 2x)logx.

 

(vi) (sinx + cosx)e^ax

Solution:

Let y = (sinx + cosx)e^ax

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}${(sinx + cosx)e^ax }

= (sinx + cosx)$\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}}} \right)}}{{{\rm{dx}}}}$ + e^ax.$\frac{{\rm{d}}}{{{\rm{dx}}}}$(sinx + cosx)

= (sinx + cosx).e^ax.a.1 + e^ax.(cosx – sinx)

 = a.(sinx + cosx).e^ax + (cosx – sinx).e^ax.

 

5. Calculate the derivative of:

(i)$\frac{{{\rm{logx}}}}{{{\rm{sinx}}}}$

Solution:

Let y = $\frac{{{\rm{logx}}}}{{{\rm{sinx}}}}$

Differentiating both sides w.r.t. x

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

= $\frac{{{\rm{sinx}}.\frac{1}{{\rm{x}}} - {\rm{logx}}.{\rm{cosx}}}}{{{{\sin }^2}{\rm{x}}}}$ = $\frac{{{\rm{sinx}} - {\rm{xcosx}}.{\rm{logx}}}}{{{\rm{x}}.{{\sin }^2}{\rm{x}}}}$.

 

(ii)$\frac{{{\rm{log}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{{\rm{e}}^{{\rm{px}}}}}}$

Solution:

Let y = $\frac{{{\rm{log}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{{\rm{e}}^{{\rm{px}}}}}}$

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{\rm{log}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{{\rm{e}}^{{\rm{px}}}}}}} \right)$ = $\frac{{{{\rm{e}}^{{\rm{px}}}}\frac{{{\rm{d}}\left( {\log \left( {{\rm{ax}} + {\rm{b}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{\rm{dx}}}} - \log \left( {{\rm{ax}} + {\rm{b}}} \right).\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{px}}}}} \right)}}{{{\rm{d}}\left( {{\rm{px}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{px}}} \right)}}{{{\rm{dx}}}}{\rm{\: \: \: \: \: }}}}{{{{\left( {{{\rm{e}}^{{\rm{px}}}}} \right)}^2}}}$.

= $\frac{{{{\rm{e}}^{{\rm{px}}}}.\frac{1}{{{\rm{ax}} + {\rm{b}}}}.{\rm{a}}.1 - \log \left( {{\rm{ax}} + {\rm{b}}} \right).{{\rm{e}}^{{\rm{px}}}}.{\rm{p}}.1}}{{{{\left( {{{\rm{e}}^{{\rm{px}}}}} \right)}^2}}}$ = $\frac{{{{\rm{e}}^{{\rm{px}}}}\left\{ {\frac{{\rm{a}}}{{{\rm{ax}} + {\rm{b}}}} - {\rm{plog}}\left( {{\rm{ax}} + {\rm{b}}} \right)} \right\}}}{{{{\left( {{{\rm{e}}^{{\rm{px}}}}} \right)}^2}}}$. = $\frac{{{\rm{a}} - {\rm{p}}.\left( {{\rm{ax}} - {\rm{b}}} \right).\log \left( {{\rm{ax}} + {\rm{b}}} \right)}}{{\left( {{\rm{ax}} + {\rm{b}}} \right).{{\rm{e}}^{{\rm{px}}}}}}$.

 

(iii)$\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{{\rm{cosbx}}}}$

Solution:

Let y = $\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{{\rm{cosbx}}}}$

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{{\rm{cosbx}}}}} \right)$ = $\frac{{{\rm{cosbx}}.\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}}} \right)}}{{{\rm{dx}}}} - {{\rm{e}}^{{\rm{ax}}}}.\frac{{{\rm{d}}\left( {{\rm{cosbx}}} \right)}}{{{\rm{d}}\left( {{\rm{bx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{bx}}} \right)}}{{{\rm{dx}}}}{\rm{\: \: \: \: \: }}}}{{{{\cos }^2}{\rm{bx}}}}$.

= $\frac{{{\rm{cos}}.{\rm{bx}}.{{\rm{e}}^{{\rm{ax}}}}.{\rm{a}}.1 - {{\rm{e}}^{{\rm{ax}}}}.\left( { - {\rm{sinbx}}} \right).{\rm{b}}.1}}{{{{\cos }^2}{\rm{bx}}}}$ = $\frac{{\left( {{\rm{acosbx}} + {\rm{bsinbx}}} \right).{{\rm{e}}^{{\rm{ax}}}}}}{{{{\cos }^2}{\rm{bx}}}}$.

 

(iv)$\frac{{{\rm{sinax}}}}{{1 + {\rm{logx}}}}$

Solution:

Let y = $\frac{{{\rm{sinax}}}}{{1 + {\rm{logx}}}}$

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{\rm{sinax}}}}{{1 + {\rm{logx}}}}} \right)$ = $\frac{{\left( {1 + {\rm{logx}}} \right).\frac{{{\rm{d}}\left( {{\rm{sinax}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}}} \right)}}{{{\rm{dx}}}} - {\rm{sinax}}.\frac{{{\rm{d}}\left( {1 + {\rm{logx}}} \right)}}{{{\rm{d}}\left( {\rm{x}} \right)}}{\rm{\: \: \: \: \: }}}}{{{{\cos }^2}{\rm{bx}}}}$.

= $\frac{{\left( {1 + {\rm{logx}}} \right).{\rm{cosax}}.{\rm{a}}.1 - {\rm{sinax}}.\frac{1}{{\rm{x}}}}}{{{{\left( {1 + {\rm{logx}}} \right)}^2}}}$ = $\frac{{{\rm{ax}}.{\rm{cosax}}\left( {1 + {\rm{logx}}} \right) - {\rm{sinax}}}}{{{\rm{x}}{{\left( {1 + {\rm{logx}}} \right)}^2}}}$.

 

(v)$\frac{{{\rm{logx}}}}{{{{\rm{a}}^2} + {{\rm{x}}^2}}}$

Solution:

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

Differentiating both sides w.r.t. x

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

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

 

(vi)$\frac{{{{\rm{x}}^{\rm{n}}}}}{{{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}}}$

Solution:

Let y = $\frac{{{{\rm{x}}^{\rm{n}}}}}{{{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}}}$

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{{\rm{x}}^{\rm{n}}}}}{{{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}}}} \right)$ = $\frac{{{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}.\frac{{{\rm{d}}{{\rm{x}}^{\rm{n}}}}}{{{\rm{dx}}}} - {{\rm{x}}^{\rm{n}}}\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{\rm{dx}}}}{\rm{\: \: \: \: \: \: }}}}{{{{\left( {{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}} \right)}^2}}}$.

= $\frac{{{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}.{\rm{n}}{{\rm{x}}^{{\rm{n}} - 1}} - {{\rm{x}}^{\rm{n}}}.{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}.{\rm{a}}.1}}{{{{\left( {{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}} \right)}^2}}}$ = $\frac{{{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}.\left( {{\rm{n}} - {\rm{ax}}} \right){{\rm{x}}^{{\rm{n}} - 1}}}}{{{{\left( {{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}} \right)}^2}}}$ = $\frac{{\left( {{\rm{n}} - {\rm{ax}}} \right){{\rm{x}}^{{\rm{n}} - 1}}}}{{{{\rm{e}}^{{\rm{ax}} + {\rm{b}}}}}}$.

 

6. Find the derivative when,

(i)xy=log(x² + y²)

Solution:

xy = log(x² + y²)

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

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

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

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

Or, ${\rm{x}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y = $\frac{1}{{{{\rm{x}}^2} + {{\rm{y}}^2}}}\left\{ {2{\rm{x}} + 2{\rm{y}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right\}$.

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

Or, x.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ – $\frac{{2{\rm{y}}}}{{{{\rm{x}}^2} + {{\rm{y}}^2}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}}}}{{{{\rm{x}}^2} + {{\rm{y}}^2}}} - {\rm{y}}$.

Or, $\frac{{{{\rm{x}}^3} + {\rm{x}}{{\rm{y}}^2} - 2{\rm{y}}}}{{{{\rm{x}}^2} + {{\rm{y}}^2}}}$.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}} - {{\rm{x}}^2}{\rm{y}} - {{\rm{y}}^3}}}{{{{\rm{x}}^2} + {{\rm{y}}^2}}}$.

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}} - {{\rm{x}}^2}{\rm{y}} - {{\rm{y}}^3}}}{{{{\rm{x}}^3} + {\rm{x}}{{\rm{y}}^2} - 2{\rm{y}}}}{\rm{\: \: }}$

 

(ii)x² + y² = log(x + y)

Solution:

x² + y² = log(x + y)

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

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x² + y²) = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left\{ {{\rm{log}}\left( {{\rm{x\: }} + {\rm{\: y}}} \right)} \right\}$.

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

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

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

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

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

Or, $\frac{{2{\rm{xy}} + 2{{\rm{y}}^2} - 1}}{{{\rm{x}} + {\rm{y}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{1 - 2{{\rm{x}}^2} - 2{\rm{xy}}}}{{{\rm{x}} + {\rm{y}}}}{\rm{\: }}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{1 - 2{{\rm{x}}^2} - 2{\rm{xy}}}}{{2{\rm{xy}} + 2{{\rm{y}}^2} - 1}}$.

 

(iii)e^xy = xy

Solution:

e^xy = xy

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

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(e^xy) = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{xy}}} \right)$.

Or, $\frac{{{\rm{d}}\left( {{{\rm{e}}^{{\rm{xy}}}}} \right)}}{{{\rm{d}}\left( {{\rm{xy}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{xy}}} \right)}}{{{\rm{dx}}}} = \frac{1}{{{\rm{dx}}}}.\left( {{\rm{xy}}} \right)$

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

Or,e^xy x. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + e^xy.y.1 = x$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y.1

Or, e^xy.x $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ – x. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = y – ye^xy

Or, x(e^xy – 1)$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = –y(e^xy – 1).

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

(iv)x = e^cos2t

Solution:

x = e^cos2t

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

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = $\frac{{{\rm{d}}{{\rm{e}}^{{\rm{cos}}2{\rm{t}}}}}}{{{\rm{dt}}}}$ = $\frac{{{\rm{d}}{{\rm{e}}^{{\rm{cos}}2{\rm{t}}}}}}{{{\rm{d}}\left( {{\rm{cos}}2{\rm{t}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cos}}2{\rm{t}}} \right)}}{{{\rm{dt}}}}$

= e^cos2t. $\frac{{{\rm{d}}\left( {{\rm{cos}}2{\rm{t}}} \right)}}{{{\rm{d}}\left( {2{\rm{t}}} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{t}}} \right)}}{{{\rm{dt}}}}$ = e^cos2t.(–sin2t).2 = –2.sin2t.e

Again, y = e^sin2t

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

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

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

= e^sin2t.cos 2t.2 = 2cos2t.e^sin2t.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}}{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}} = \frac{{2{\rm{cos}}2{\rm{t}}.{{\rm{e}}^{{\rm{sin}}2{\rm{t}}}}}}{{ - 2{\rm{sin}}2{\rm{t}}.{{\rm{e}}^{{\rm{cos}}2{\rm{t}}}}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = –e^{sin2t – cos2t} . cos2t.

 

(v)x =cos(log t)

Solution:

x =cos(log t)

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

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = $\frac{{{\rm{d}}\left( {\log \left( {{\rm{cost}}} \right)} \right)}}{{{\rm{dt}}}}$ = $\frac{{{\rm{d}}\left( {\log \left( {{\rm{cost}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{cost}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cost}}} \right)}}{{{\rm{dt}}}}$

= – sin(log t) $\frac{1}{{\rm{t}}}$ = $ - \frac{1}{{\rm{t}}}$.sin(log t)

Again, y = log(cos t)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = $\frac{{{\rm{d}}\left( {\log \left( {{\rm{cost}}} \right)} \right)}}{{{\rm{dt}}}}.\frac{{{\rm{d}}\left( {\log \left( {{\rm{cost}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{cost}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cost}}} \right)}}{{{\rm{dt}}}}$ = $\frac{1}{{{\rm{cost}}}}.\left( { - {\rm{sint}}} \right)$ = – tant.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}}{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}} = \frac{{ - {\rm{tant}}}}{{ - \frac{1}{{\rm{t}}}{\rm{sin}}\left( {{\rm{logt}}} \right)}}$ = $\frac{{{\rm{t}}.{\rm{tan}}.{\rm{t}}}}{{\sin \left( {{\rm{logt}}} \right)}}$.

 

(vi)x =log t + sin t

Solution:

x =log t + sin t

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

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = $\frac{{{\rm{d}}\left( {\log {\rm{t}} + {\rm{sint}}} \right)}}{{{\rm{dt}}}}$ = $\frac{1}{{\rm{t}}} + {\rm{cost}}$ = $\frac{{1 + {\rm{t}}.{\rm{cost}}}}{{\rm{t}}}$

Again, y = e^t + cos.t

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = $\frac{{{\rm{d}}\left( {{{\rm{e}}^{\rm{t}}}{\rm{\: }} + {\rm{\: cos}}.{\rm{t}}} \right)}}{{{\rm{dt}}}}$ = e^t – sin.t

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}}{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}} = \frac{{{{\rm{e}}^{\rm{t}}} - {\rm{sint}}}}{{\frac{{1 + {\rm{t}}.{\rm{cot}}.{\rm{t}}}}{{\rm{t}}}}}$ = $\frac{{{\rm{t}}\left( {{{\rm{e}}^{\rm{t}}} - {\rm{sint}}} \right)}}{{1 + {\rm{t}}.{\rm{cot}}.{\rm{t}}}}$