Derivative Exercise: 17.2 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 17.2

1. Find From the first principle, the differential coefficients of:

(i) sin4x

Solution:

Let y = sin4x

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively.Then,

Or, y + $\Delta $y = sin4 (x + $\Delta $x)

Or, $\Delta $y = sin(4x + 4$\Delta $x) – y

= sin(4x + 4$\Delta $x) – sin4x

= 2cos $\frac{{4{\rm{x}} + 4\Delta {\rm{x}} + 4{\rm{x}}}}{2}$. Sin $\frac{{4{\rm{x}} + 4\Delta {\rm{x}} - 4{\rm{x}}}}{2}$

= 2cos(4x + 2$\Delta $x)sin (2$\Delta $x)

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{2\cos \left( {4{\rm{x}} + 2\Delta {\rm{x}}} \right)\sin \left( {2\Delta {\rm{x}}} \right)}}{{\Delta {\rm{x}}}}$

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 2cos(4x + 2$\Delta {\rm{x}}$) $\frac{{{\rm{sin}}2\Delta {\rm{x}}}}{{2\Delta {\rm{x}}}}$ * 2.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 2cos(4x + 0).1 * 2 = 4cos4x.

 

(ii)cos(ax – b)

Solution:

Let y = cos(ax – b)

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively.Then,

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

Or, $\Delta $y = cos(ax + a$\Delta $x – b) – y

= cos(ax + a$\Delta $x – b) – cos(ax – b).

= 2sin $\frac{{{\rm{ax}} + {\rm{a}}\Delta {\rm{x}} - {\rm{b}} + {\rm{ax}} - {\rm{b}}}}{2}$. Sin $\frac{{{\rm{ax}} - {\rm{b}} - {\rm{ax}} - {\rm{a}}\Delta {\rm{x}} + {\rm{b}}}}{2}$

= 2sin$\left( {{\rm{ax}} - {\rm{b}} + \frac{{{\rm{a}}\Delta {\rm{x}}}}{2}} \right)$ sin $\frac{{ - {\rm{a}}\Delta {\rm{x}}}}{2}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{ - {\rm{\: }}2{\rm{sin\: }}\left( {{\rm{ax}} - {\rm{b}} + \frac{{{\rm{a}}\Delta {\rm{x}}}}{2}} \right).\sin \frac{{{\rm{a}}\Delta {\rm{x}}}}{2}}}{{\Delta {\rm{x}}}}$

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 $ - {\rm{\: }}2{\rm{sin\: }}\left( {{\rm{ax}} - {\rm{b}} + \frac{{{\rm{a}}\Delta {\rm{x}}}}{2}} \right)$. $\frac{{\sin \frac{{{\rm{a}}\Delta {\rm{x}}}}{2}}}{{\frac{{{\rm{a}}\Delta {\rm{x}}}}{2}}}$.$\frac{{\rm{a}}}{2}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = –2sin(ax – b + 0).1.$\frac{{\rm{a}}}{2}$ = –asin(ax – b).

 

(iii)tan(3x – 4)

Solution:

Let y = tan(3x – 4)

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively.Then,

Or, y + $\Delta $y = tan{3(x + $\Delta $x) – 4}

Or, $\Delta $y = tan (3x + 3$\Delta $x – 4 – y)

= tan (3x + 3$\Delta $x – 4 – y) – tan(3x – 4)

= $\frac{{\sin \left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4} \right)}}{{\cos \left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4} \right)}} - \frac{{\sin \left( {3{\rm{x}} - 4} \right)}}{{{\rm{cos}}\left( {3{\rm{x}} - 4} \right)}}$

= $\frac{{\sin \left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4} \right).\cos \left( {3{\rm{x}} - 4} \right) - \cos \left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4} \right).\sin \left( {3{\rm{x}} - 4} \right)}}{{{\rm{cos}}\left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4} \right)\cos \left( {3{\rm{x}} - 4} \right)}}$.

= $\frac{{\sin \left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4 - 3{\rm{x}} + 4} \right)}}{{\cos \left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4} \right).\cos \left( {3{\rm{x}} - 4} \right)}}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{{\rm{sin}}3\Delta {\rm{x}}}}{{\Delta {\rm{xcos}}\left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4} \right){\rm{cos}}\left( {3{\rm{x}} - 4} \right)}}$

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 $\frac{{{\rm{sin}}3\Delta {\rm{x}}}}{{3\Delta {\rm{x}}}}{\rm{*}}3{\rm{*}}\frac{1}{{\cos \left( {3{\rm{x}} + 3\Delta {\rm{x}} - 4} \right)\cos \left( {3{\rm{x}} - 4} \right)}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 1.3. $\frac{1}{{\cos \left( {3{\rm{x}} + 0 - 4} \right).\cos \left( {3{\rm{x}} - 4} \right)}}$ = 3sec²(3x – 4).

 

(iv)sin $\frac{{3{\rm{x}}}}{2}$

Solution:

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

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively.Then,

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

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

= 2cos $\frac{{3{\rm{x}} + 3\Delta {\rm{x}} + 3{\rm{x}}}}{4}$. Sin $\frac{{3{\rm{x}} + 3\Delta {\rm{x}} - 3{\rm{x}}}}{4}$

= 2cos $\left( {\frac{{3{\rm{x}}}}{2} + \frac{{3\Delta {\rm{x}}}}{4}} \right)$.sin $\frac{{3\Delta {\rm{x}}}}{4}$.

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{2\cos \left( {\frac{{3{\rm{x}}}}{2} + \frac{{3\Delta {\rm{x}}}}{4}} \right)\sin \frac{{3\Delta {\rm{x}}}}{4}{\rm{\: }}}}{{\Delta {\rm{x}}}}$

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 $2\cos \left( {\frac{{3{\rm{x}}}}{2} + \frac{{3\Delta {\rm{x}}}}{2}} \right)\left( {\frac{{\sin \frac{{3\Delta {\rm{x}}}}{4}{\rm{\: }}}}{{\frac{{3\Delta {\rm{x}}}}{4}}}} \right)$ = $\frac{3}{4}{\rm{\: \: }}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = 2cos $\left( {\frac{{3{\rm{x}}}}{2} + 0} \right)$.1.$\frac{3}{4}$ = $\frac{3}{2}$cos $\frac{{3{\rm{x}}}}{2}$.

 

(v)tan $\frac{{5{\rm{x}}}}{3}$

Solution:

Let y = tan $\frac{{5{\rm{x}}}}{3}$

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively. Then,

Or, y + $\Delta $y = tan$\frac{{5\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}{3}$

Or, $\Delta $y = tan $\frac{{5\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}{3} - {\rm{y}}$ = tan $\frac{{5\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}{3} - \tan \frac{{5{\rm{x}}}}{3}$

= $\frac{{\sin \frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}}}{{\cos \frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}}} - \frac{{\sin \frac{{5{\rm{x}}}}{3}}}{{\cos \frac{{5{\rm{x}}}}{3}}}$

= $\frac{{\sin \frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}.\cos \frac{{5{\rm{x}}}}{3} - \cos \frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}.\sin \frac{{5{\rm{x}}}}{3}}}{{\cos \frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}.\cos \frac{{5{\rm{x}}}}{3}}}$

= $\frac{{\sin \left( {\frac{{5{\rm{x}} + 5\Delta }}{3} - \frac{{5{\rm{x}}}}{3}} \right)}}{{\cos \frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}.\cos \frac{{5{\rm{x}}}}{3}}}$

= $\frac{{\sin \frac{{5\Delta {\rm{x}}}}{3}}}{{\cos \frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}.\cos \frac{{5{\rm{x}}}}{3}}}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = $\frac{{\sin \frac{{5\Delta {\rm{x}}}}{3}}}{{\Delta {\rm{xcos}}\frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}.\cos \frac{{5{\rm{x}}}}{3}}}$

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 $\frac{{\sin \frac{{5\Delta {\rm{x}}}}{3}}}{{\frac{{5\Delta {\rm{x}}}}{3}}}.\frac{5}{3}.\frac{1}{{\cos \frac{{5{\rm{x}} + 5\Delta {\rm{x}}}}{3}{\rm{\: }}.\cos \frac{{5{\rm{x}}}}{3}}}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$= 1.$\frac{5}{3}$.$\frac{1}{{\cos \frac{{5{\rm{x}}}}{3}.\cos \frac{{5{\rm{x}}}}{3}}}$ = $\frac{5}{3}.{\sec ^2}\frac{{5{\rm{x}}}}{3}$

 

(vi)cos²x 

Solution:

Let y = cos²x = $\frac{1}{2}\left( {2{{\cos }^2}{\rm{x}}} \right)$ = $\frac{1}{2}\left( {1 + {\rm{cos}}2{\rm{x}}} \right)$

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively. Then,

Or, y + $\Delta $y = $\frac{1}{2}\left[ {1 + {\rm{cos}}2\left( {{\rm{x}} + \Delta {\rm{x}}} \right)} \right]$

Or, $\Delta $y = $\frac{1}{2}\left[ {1 + \cos \left( {2{\rm{x}} + 2\Delta {\rm{x}}} \right)} \right] - {\rm{y\: }}$= $\frac{1}{2}\left[ {1 + \cos \left( {2{\rm{x}} + 2\Delta {\rm{x}}} \right)} \right] - \frac{1}{2}\left( {1 + {\rm{cos}}2{\rm{x}}} \right)$

= $\frac{1}{2}\left[ {1 + \cos \left( {2{\rm{x}} + 2\Delta {\rm{x}}} \right) - 1 - {\rm{cos}}2{\rm{x}}} \right]$

= $\frac{1}{2}\left[ {1 + \cos \left( {2{\rm{x}} + 2\Delta {\rm{x}}} \right) - {\rm{cos}}2{\rm{x}}} \right]$.

= $\frac{1}{2}2\sin \frac{{2{\rm{x}} + 2{\rm{x}} + 2\Delta {\rm{x}}}}{2}.\sin \frac{{2{\rm{x}} - 2{\rm{x}} - 2\Delta {\rm{x}}}}{2}$

= sin (2x + $\Delta $x).sin(–$\Delta $).

= – sin(2x + 2$\Delta $x).sin$\Delta $x

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

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 – sin(2x + $\Delta $x)$\frac{{{\rm{sin}}\Delta {\rm{x}}}}{{\Delta {\rm{x}}}}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = – sin(2x + 0).1 = –sin2x.

 

(vii)sin²3x 

Solution:

Let y = sin²3x = $\frac{1}{2}\left( {2{{\sin }^2}3{\rm{x}}} \right)$ = $\frac{1}{2}\left( {1 - {\rm{cos}}6{\rm{x}}} \right)$

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively. Then,

Or, y + $\Delta $y = $\frac{1}{2}\left[ {1 - {\rm{cos}}6\left( {{\rm{x}} + \Delta {\rm{x}}} \right)} \right]$

Or, $\Delta $y = $\frac{1}{2}\left[ {1 - \cos \left( {6{\rm{x}} + 6\Delta {\rm{x}}} \right)} \right] - {\rm{y\: }}$= $\frac{1}{2}\left[ {1 - \cos \left( {6{\rm{x}} + 6\Delta {\rm{x}}} \right)} \right] - \frac{1}{2}\left[ {1 - {\rm{cos}}6{\rm{x}}} \right]$

= $\frac{1}{2}\left[ {1 - \cos \left( {6{\rm{x}} + 6\Delta {\rm{x}}} \right) - 1 + {\rm{cos}}6{\rm{x}}} \right]$

= $\frac{1}{2}\left[ {{\rm{cos}}6{\rm{x}} - {\rm{cos}}\left( {6{\rm{x}} + 6\Delta {\rm{x}}} \right)} \right]$.

= $\frac{1}{2}2\sin \frac{{6{\rm{x}} + 6{\rm{x}} + 6\Delta {\rm{x}}}}{2}.\sin \frac{{6{\rm{x}} + 6\Delta {\rm{x}} - 6{\rm{x}}}}{2}$

= sin (6x + 3$\Delta $x)sin3$\Delta $x.

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ = sin (6x + 3$\Delta $x). $\frac{{{\rm{sin}}3\Delta {\rm{x}}}}{{3\Delta {\rm{x}}}}$

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 sin(6x + 3$\Delta $x) $\frac{{{\rm{sin}}3\Delta {\rm{x}}}}{{3\Delta {\rm{x}}}}$.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$= sin(6x + 0).1.3 = 3sin6x.

 

(viii) $\sqrt {{\rm{sin}}2{\rm{x}}} $

Solution:

Let y =  $\sqrt {{\rm{sin}}2{\rm{x}}} $

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively. Then,

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

Or, $\Delta $y = $\sqrt {{\rm{sin}}2\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  - {\rm{y\: }}$= $\sqrt {{\rm{sin}}2\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  - \sqrt {{\rm{sin}}2{\rm{x}}} $

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

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

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

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

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

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 $\frac{{2\cos \left( {2{\rm{x}} + \Delta {\rm{x}}} \right)}}{{\sqrt {\sin \left( {2{\rm{x}} + 2\Delta {\rm{x}}} \right)}  + \sqrt {{\rm{sin}}2{\rm{x}}} }}.\frac{{{\rm{sin}}\Delta {\rm{x}}}}{{\Delta {\rm{x}}}}$

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

 

(ix)$\sqrt {{\rm{secx}}} $

Solution:

Let y =  $\sqrt {{\rm{secx}}} $

Let $\Delta $x and $\Delta $y ne the small increments in x and y respectively. Then,

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

Or, $\Delta $y = $\sqrt {{\rm{sec}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  - {\rm{y\: }}$= $\sqrt {{\rm{sec}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  - \sqrt {{\rm{secx}}} $

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

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

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

= $\frac{{{\rm{cosx}} - {\rm{cos}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right)}}{{\cos \left( {{\rm{x}} + \Delta {\rm{x}}} \right).{\rm{cosx}}\left( {\sqrt {\sec \left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  + \sqrt {{\rm{secx}}} } \right)}}$

= $\frac{{2\sin \frac{{{\rm{x}} + {\rm{x}} + \Delta {\rm{x}}}}{2}.\sin \frac{{{\rm{x}} + \Delta {\rm{x}} - {\rm{x}}}}{2}}}{{\cos \left( {{\rm{x}} + \Delta {\rm{x}}} \right)\cos \left( {\sqrt {\sec \left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  + \sqrt {{\rm{secx}}} } \right)}}$

= $\frac{{2\sin \left( {{\rm{x}} + \frac{{\Delta {\rm{x}}}}{2}} \right).\sin \frac{{\Delta {\rm{x}}}}{2}}}{{\cos \left( {{\rm{x}} + \Delta {\rm{x}}} \right).{\rm{cosx}}\left( {\sqrt {\sec \left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  + \sqrt {{\rm{secx}}} } \right)}}$

Or, $\frac{{\Delta {\rm{y}}}}{{\Delta {\rm{x}}}}$ =$\frac{{2\sin \left( {{\rm{x}} + \frac{{\Delta {\rm{x}}}}{2}} \right).\sin \frac{{\Delta {\rm{x}}}}{2}}}{{\Delta {\rm{xcos}}\left( {{\rm{x}} + \Delta {\rm{x}}} \right).{\rm{cosx}}\left( {\sqrt {\sec \left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  + \sqrt {{\rm{secx}}} } \right)}}$

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 $\frac{{2\sin \left( {{\rm{x}} + \frac{{\Delta {\rm{x}}}}{2}} \right)}}{{\cos \left( {{\rm{x}} + \Delta {\rm{x}}} \right).{\rm{cosx}}\left( {\sqrt {\sec \left( {{\rm{x}} + \Delta {\rm{x}}} \right)}  + \sqrt {{\rm{secx}}} } \right)}}.\frac{{\sin \frac{{\Delta {\rm{x}}}}{2}}}{{\frac{{\Delta {\rm{x}}}}{2}}}.\frac{1}{2}$

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{sinx}}}}{{{\rm{cosx}}.{\rm{cosx}}\left( {\sqrt {{\rm{secx}}}  + \sqrt {{\rm{secx}}} } \right)}}1.\frac{1}{2}{\rm{\: }}$ = $\frac{{{\rm{secx}}.{\rm{tanx}}}}{{2\sqrt {{\rm{secx}}} }}$ = $\frac{1}{2}\sqrt {{\rm{secx}}} $.tanx.

 

2.Find the derivative of:

(i)sin(4x – 5)

Solution:

Let y = sin(4x – 5)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$[sin(4x – 5)]

= $\frac{{{\rm{d}}\left[ {\sin \left( {4{\rm{x}} - 5} \right)} \right]}}{{{\rm{d}}\left( {4{\rm{x}} - 5} \right)}}.\frac{{{\rm{d}}\left( {4{\rm{x}} - 5} \right)}}{{{\rm{dx}}}}$

= cos(4x – 5).4.1 = 4cos(4x – 5).

 

(ii) cos(ax + b)

Solution:

Let y = cos(ax + b)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$cos (ax + b)

= $\frac{{{\rm{dcos}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{\rm{d}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{\rm{dx}}}}$

= – sin(ax + b).a.1 = –asin(ax + b).

 

(iii)tan(5x² + 6)

Solution:

Let y = tan(5x² + 6)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$tan(5x² + 6)

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

= sec²(5x² + 6).5.2x = 10x.sec²(5x² + 6).

 

(iv)cot$\sqrt {\rm{x}} $

Solution:

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

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

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

= $\frac{{{\rm{d}}\left( {{\rm{cot}}\sqrt {\rm{x}} } \right)}}{{{\rm{d}}\left( {\sqrt {\rm{x}} } \right)}}.\frac{{{\rm{d}}\left( {\sqrt {\rm{x}} } \right)}}{{{\rm{dx}}}}$

= –cosec²x $\sqrt {\rm{x}} $. $\frac{1}{{2\sqrt {\rm{x}} }}$ = $ - \frac{1}{{2\sqrt {\rm{x}} }}$ cosec²$\sqrt {\rm{x}} $.

 

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

Solution:

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

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$ (sec $\frac{1}{{\rm{x}}}$)

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

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

 

(vi)cosec $\frac{{{\rm{a}}{{\rm{x}}^2}}}{{\rm{b}}}$

Solution:

Let y = cosec $\frac{{{\rm{a}}{{\rm{x}}^2}}}{{\rm{b}}}$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$ (cosex $\frac{{{\rm{a}}{{\rm{x}}^2}}}{{\rm{b}}}$)

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

= – cosec $\frac{{{\rm{a}}{{\rm{x}}^2}}}{{\rm{b}}}$.cot $\frac{{{\rm{a}}{{\rm{x}}^2}}}{{\rm{b}}}.\frac{{\rm{a}}}{{\rm{b}}}$.2x = –$\frac{{2{\rm{ax}}}}{{\rm{b}}}$cosec $\frac{{{\rm{a}}{{\rm{x}}^2}}}{{\rm{b}}}$.cot $\frac{{{\rm{a}}{{\rm{x}}^2}}}{{\rm{b}}}$.

 

(vii)sin⁵(ax² – c)

Solution:

Let y = sin⁵(ax² – c)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$ [sin⁵(ax² – c)]

= $\frac{{{\rm{d}}[{{\sin }^5}\left( {{\rm{a}}{{\rm{x}}^2} - {\rm{c}}} \right)]}}{{{\rm{d}}\left( {\sin \left( {{\rm{a}}{{\rm{x}}^2} - {\rm{c}}} \right)]} \right)}}.\frac{{{\rm{d}}\left( {{{\sin }^5}\left( {{\rm{a}}{{\rm{x}}^2} - {\rm{c}}} \right)]} \right)}}{{{\rm{d}}\left( {{\rm{a}}{{\rm{x}}^2} - {\rm{c}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{a}}{{\rm{x}}^2} - {\rm{c}}} \right)}}{{{\rm{dx}}}}$.

= 5sin⁴(ax² – c)cos(ax² – c).a.2x

= 10ax.sin⁴x(ax² – c)cos(ax² – c).

 

(viii)cos³(2ax – 3b)

Solution:

Let y = cos³(2ax – 3b)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$ [cos³(2ax – 3b)]

= $\frac{{{\rm{d}}[{{\cos }^3}\left( {2{\rm{ax\: }}--{\rm{\: }}3{\rm{b}}} \right)]}}{{{\rm{d}}\left( {{\rm{cos}}\left( {2{\rm{ax\: }}--{\rm{\: }}3{\rm{b}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cos}}\left( {2{\rm{ax\: }}--{\rm{\: }}3{\rm{b}}} \right)} \right)}}{{{\rm{d}}\left( {2{\rm{ax}} - 3{\rm{b}}} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{ax}} - 3{\rm{b}}} \right)}}{{{\rm{dx}}}}$.

= 3cos²(2ax – 3b)(–sin(2ax – 3b)).2a.1

= –6a.cos²(2ax – 3b).sin(2ax – 3b).

 

(ix)a$\sqrt {\tan \left( {5{\rm{x}} - 7} \right)} $

Solution:

Let y = a$\sqrt {\tan \left( {5{\rm{x}} - 7} \right)} $ = atan^1/2(5x – 7)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = ${\rm{a}}.\frac{{\rm{d}}}{{{\rm{dx}}}}$ (tan^1/2(5x – 7))

= $\frac{{{\rm{d}}[\left( {{{\tan }^{\frac{1}{2}}}\left( {5{\rm{x\: }}--{\rm{\: }}7} \right)} \right)}}{{{\rm{d}}\left( {{\rm{tan}}\left( {5{\rm{x\: }}--{\rm{\: }}7} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{tan}}\left( {5{\rm{x\: }}--{\rm{\: }}7} \right)} \right)}}{{{\rm{d}}\left( {5{\rm{x}} - 7} \right)}}.\frac{{{\rm{d}}\left( {5{\rm{x}} - 7} \right)}}{{{\rm{dx}}}}$.

= a.$\frac{1}{2}$.tan^–1/2(5x – 7).sec²(5x – 7).5.1

= $\frac{{5{\rm{ase}}{{\rm{c}}^2}\left( {5{\rm{x}} - 7} \right)}}{{2\sqrt {\tan \left( {5{\rm{x}} - 7} \right)} }}$.

 

(x) sec⁵$\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)$

Solution:

Let y = sec⁵$\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\sec }^5}\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)} \right)$

= $\frac{{{\rm{d}}\left( {{{\sec }^5}\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)} \right)}}{{{\rm{d}}\left( {\sec \left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {{{\sec }^5}\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)} \right)}}{{{\rm{d}}\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)}}{{{\rm{dx}}}}$.

= 5.sec⁴$\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)$.sec $\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)$.tan$\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)$.$\frac{{{\rm{a}}.1}}{{\rm{c}}}$

= $\frac{{5{\rm{a}}}}{{\rm{c}}}$ sec⁵$\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)$.tan$\left( {\frac{{{\rm{ax}} + {\rm{b}}}}{{\rm{c}}}} \right)$.

 

(xi)cosecⁿ$\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)$
Solution:

Let y = cosecⁿ$\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{cose}}{{\rm{c}}^{\rm{n}}}\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)} \right)$

= $\frac{{{\rm{d}}\left( {{\rm{cose}}{{\rm{c}}^{\rm{n}}}\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{cosec}}\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cosec}}\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)} \right)}}{{{\rm{d}}\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}{\rm{\: }}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)}}{{{\rm{dx}}}}$.

= n.cosec^n–1.$\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)$.$\left( { - {\rm{cosex}}\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)} \right).\left( {\cot \left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)} \right).\frac{{\rm{p}}}{{\rm{r}}}.2{\rm{x}}$

= $ - \frac{{2{\rm{npx}}}}{{\rm{r}}}$.cosecⁿ$\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)$.cot$\left( {\frac{{{\rm{p}}{{\rm{x}}^2} - {\rm{q}}}}{{\rm{r}}}} \right)$.

 

3. Find the differential coefficient of 

(i)tan(cos 5x)

Solution:

Let y = tan(cos 5x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan(cos 5x))

= $\frac{{{\rm{d}}\left( {\tan \left( {{\rm{cos}}5{\rm{x}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{cos}}5{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cos}}5{\rm{x}}} \right)}}{{{\rm{d}}\left( {5{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {5{\rm{x}}} \right)}}{{{\rm{dx}}}}$

= sec²(cos 5x).(–sin5x).5.1 = – 5sec²(cos 5x)sin5x.

 

(ii) cos(sin(3x² + 2))

Solution:

Let y = cos(sin(3x² + 2))

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

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

= $\frac{{{\rm{d}}\left\{ {{\rm{cos}}\left( {{\rm{sin}}\left( {3{{\rm{x}}^2}{\rm{\: }} + {\rm{\: }}2} \right)} \right)} \right\}}}{{{\rm{d}}\left( {{\rm{sin}}3{{\rm{x}}^2} + 2} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sin}}3{{\rm{x}}^2} + 2} \right)}}{{{\rm{d}}\left( {3{{\rm{x}}^2} + 2} \right)}}.\frac{{{\rm{d}}\left( {3{{\rm{x}}^2} + 2} \right)}}{{{\rm{dx}}}}$

= – sin(sin(3x² + 2)).cos(3x² + 2).3.2x

= – 6x sin(sin3x² + 2).cos(3x² + 2).

 

(iii)sin(tan(ax + b))

Solution:

Let y = sin(tan(ax + b))

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sin(tan(ax + b)))

= $\frac{{{\rm{d}}\left( {{\rm{sin}}\left( {{\rm{tan}}\left( {{\rm{ax\: }} + {\rm{\: b}}} \right)} \right)} \right)}}{{{\rm{d}}\left( {\tan \left( {{\rm{ax}} + {\rm{b}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {\tan \left( {{\rm{ax}} + {\rm{b}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{\rm{dx}}}}$

= cos.(tan(ax + b)).sec²(ax + b).a.1

= a.cos(tan(ax + b)).sec²(ax + b).

 

(iv)sec²(tan $\sqrt {\rm{x}} $)

Solution:

Let y = sec²(tan $\sqrt {\rm{x}} $)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sec²(tan $\sqrt {\rm{x}} $))

= $\frac{{{\rm{d}}\left( {{{\sec }^2}\left( {\tan \sqrt {\rm{x}} } \right)} \right)}}{{{\rm{d}}\left( {{\rm{sec}}.\left( {{\rm{tan}}\sqrt {\rm{x}} } \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sec}}.\left( {{\rm{tan}}\sqrt {\rm{x}} } \right)} \right)}}{{{\rm{d}}\left( {{\rm{tan}}\sqrt {\rm{x}} } \right)}}.\frac{{{\rm{d}}\left( {{\rm{tan}}\sqrt {\rm{x}} )} \right)}}{{{\rm{d}}\left( {\sqrt {\rm{x}} } \right)}}.\frac{{{\rm{d}}\sqrt {\rm{x}} }}{{{\rm{dx}}}}$

= 2sec.(tan $\sqrt {\rm{x}} $).sec(tan$\sqrt {\rm{x}} $).tan(tan $\sqrt {\rm{x}} $).sec²$\left( {\sqrt {\rm{x}} } \right)$.$\frac{1}{{2\sqrt {\rm{x}} }}$

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

 

(v) tan⁵(sin(px – q))

Solution:

Let y = tan⁵(sin(px – q))

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan⁵(sin(px – q))

= $\frac{{{\rm{d}}\left( {{{\tan }^5}\left( {{\rm{sin}}\left( {{\rm{px\: }}--{\rm{\: q}}} \right)} \right)} \right)}}{{{\rm{d}}\left( {{\rm{tan}}({\rm{sin}}\left( {{\rm{px}} - {\rm{q}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{tan}}({\rm{sin}}\left( {{\rm{px}} - {\rm{q}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{sin}}\left( {{\rm{px}} - {\rm{q}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sin}}\left( {{\rm{px}} - {\rm{q}}} \right))} \right)}}{{{\rm{d}}\left( {{\rm{px}} - {\rm{q}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{px}} - {\rm{q}}} \right)}}{{{\rm{dx}}}}$

= 5tan⁴(sin(px – q)).sec²(sin(px – q)).cos(px – q).p.1

= 5p.tan⁴(sin(px – q))sec²(sin(px – q)).cos(px – q).

 

(vi) cosec³(cot 4x)

Solution:

Let y = cosec³(cot 4x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan⁵(sin(px – q))

= $\frac{{{\rm{d}}\left( {{\rm{cose}}{{\rm{c}}^3}\left( {{\rm{cot\: }}4{\rm{x}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{cosec}}\left( {{\rm{cot\: }}4{\rm{x}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cosec}}\left( {{\rm{cot\: }}4{\rm{x}}} \right))} \right)}}{{{\rm{d}}\left( {{\rm{cot}}\left( {4{\rm{x}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cot}}\left( {4{\rm{x}}} \right)} \right)}}{{{\rm{d}}\left( {4{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {4{\rm{x}}} \right)}}{{{\rm{dx}}}}$

= 3cosec²(cot 4x).(–cosec(cot4x)).cot(cot 4x).(–cosec²4x).4.1

= 12cosec³(cot 4x).cot(cot4x).cosec²4x.

 

(vii) cot$\left( {\sqrt {{\rm{tan}}3{\rm{x}}} } \right)$

Solution:

Let y = cot$\left( {\sqrt {{\rm{tan}}3{\rm{x}}} } \right)$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(cot$\left( {\sqrt {{\rm{tan}}3{\rm{x}}} } \right)$

= $\frac{{{\rm{d}}\left( {{\rm{cot}}\left( {\sqrt {{\rm{tan}}3{\rm{x}}} } \right))} \right)}}{{{\rm{d}}\left( {\sqrt {{\rm{tan}}3{\rm{x}}} )} \right)}}.\frac{{{\rm{d}}\left( {\sqrt {{\rm{tan}}3{\rm{x}}} } \right)}}{{{\rm{d}}\left( {{\rm{tan}}3{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{tan}}3{\rm{x}}} \right)}}{{{\rm{d}}\left( {3{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {3{\rm{x}}} \right)}}{{{\rm{dx}}}}$

= –cosec²($\sqrt {{\rm{tan}}3{\rm{x}}} $).$\frac{1}{2}$(tan 3x)^–1/2.sec².3x.3.1

= $ - \frac{{3{\rm{cose}}{{\rm{c}}^2}\left( {\sqrt {{\rm{tan}}3{\rm{x}}} } \right){{\sec }^2}3{\rm{x}}}}{{2\sqrt {{\rm{tan}}3{\rm{x}}} }}{\rm{\: }}$

 

(viii) sin²(cos 6x)

Solution:

Let y = sin²(cos 6x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sin²(cos 6x))

= $\frac{{{\rm{d}}\left( {{{\sin }^2}({\rm{cos\: }}6{\rm{x}})} \right)}}{{{\rm{d}}\left( {{\rm{sin}}\left( {{\rm{cos}}6{\rm{x}}} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sin}}\left( {{\rm{cos}}6{\rm{x}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{cos}}6{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{cos}}6{\rm{x}}} \right)}}{{{\rm{d}}\left( {6{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {6{\rm{x}}} \right)}}{{{\rm{dx}}}}$

= 2sin(cos 6x).cos(cos 6x).(–sin6x).6.1

= –6{2sin(cos6x).cos(cos6x)}sin6x

= –6sin(2cos6x).sin6x.

 

4. Find the derivative of: 

(i)(x² + 3x).sin 5x

Solution:
Let y = (x² + 3x).sin 5x

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

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

= $\left( {{{\rm{x}}^2} + 3{\rm{x}}} \right)\frac{{{\rm{d}}\left( {{\rm{sin}}5{\rm{x}}} \right)}}{{{\rm{dx}}}} + {\rm{sin}}5{\rm{x}}\frac{{{\rm{d}}\left( {{{\rm{x}}^2} + 3{\rm{x}}} \right)}}{{{\rm{dx}}}}$

= (x² + 3x).cos5x.5 .1 + sin5x.(2x + 3)

= 5(x² + 3x).cos5x.5.1 + sin5x.(2x + 3).

= 5(x² + 3x).cos5x + (2x + 3)sin5x

 

(ii)x³ tan(2x³ + 3x)

Solution:
Let y = x³ tan(2x³ + 3x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x³ tan(2x³ + 3x))

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

= x³ sec²(2x³ + 3x).(6x² + 3) + tan(2x³ + 3x).3x²

= 3x³(2x² + 1)sec²(2x³ + 3x) + 3x².tan(2x³ + 3x).

 

(iii)${\rm{a}}\sqrt {\rm{x}} $cos(ax² – b)

Solution:
Let y = ${\rm{a}}\sqrt {\rm{x}} $cos(ax² – b)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(${\rm{a}}\sqrt {\rm{x}} $cos(ax² – b))

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

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

= $\frac{{\rm{a}}}{{2\sqrt {\rm{x}} }}$cos(ax² – b) – 2a²x^3/2sin(ax² – b).

 

(iv)(x + sin2x)sec3x²

Solution:
Let y = (x + sin2x)sec3x²

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$((x + sin2x)sec3x²)

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

= (x + sin2x)sec(3x²).tan(3x²)6x + sec(3x²) $\left\{ {\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right) + \frac{{{\rm{d}}\left( {{\rm{sin}}2{\rm{x}}} \right)}}{{{\rm{d}}\left( {2{\rm{x}}} \right)}}.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {2{\rm{x}}} \right)} \right\}$

= 6x(x + sin2x)sec3x².tan3x² + sec3x²(1 + cos2x.x)

=6x(x + sin2x)sec3x²tan3x² + sec3x²(1 + 2cos²x).

 

(v) ax³.cosec(p – qx)

Solution:
Let y = ax³.cosec(p – qx)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = ${\rm{a}}.\frac{{\rm{d}}}{{{\rm{dx}}}}$(x³.cosec(p – qx))

= ${\rm{a}}.\left\{ {{{\rm{x}}^3}\frac{{{\rm{d}}\left( {{\rm{cosec}}\left( {{\rm{p}} - {\rm{qx}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{p}} - {\rm{qx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{p}} - {\rm{qx}}} \right)}}{{{\rm{dx}}}} + {\rm{cosec}}\left( {{\rm{p}} - {\rm{qx}}} \right)\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^3}} \right)} \right\}{\rm{\: }}$

= a{x³ (–cosec(p – qx).cot(p – qx)(–q.1) + cosec(p – qx)3x²}

= aqx³cosec(p – qx)cot(p – qx) + 3ax²cosec(p – qx)

 

(vi)$\frac{1}{{\sqrt {\rm{x}} }}.{\rm{sin}}\sqrt {\rm{x}} $

Solution:
Let y = $\frac{1}{{\sqrt {\rm{x}} }}.{\rm{sin}}\sqrt {\rm{x}} $ = $\frac{{{\rm{sin}}\sqrt {\rm{x}} }}{{\sqrt {\rm{x}} }}$

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

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

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

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

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

 

(vii) $\frac{{{{\rm{x}}^2} - 1}}{{{\rm{cos}}4{\rm{x}}}}$

Solution:
Let y = $\frac{{{{\rm{x}}^2} - 1}}{{{\rm{cos}}4{\rm{x}}}}$

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

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

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

= $\frac{{{\rm{cos}}4{\rm{x}}.2{\rm{x}} - \left( {{{\rm{x}}^2} - 1} \right).\left( { - {\rm{sin}}4{\rm{x}}} \right).4.1}}{{{{\cos }^2}4{\rm{x}}}}$ = $\frac{{2{\rm{x}}.{\rm{cos}}4{\rm{x}} + 4\left( {{{\rm{x}}^2} - 1} \right).{\rm{sin}}4{\rm{x}}}}{{{{\cos }^2}4{\rm{x}}}}$

 

(viii)$\frac{{{\rm{secnx}}}}{{{\rm{ax}} - {\rm{b}}}}$

Solution:
Let y = $\frac{{{\rm{secnx}}}}{{{\rm{ax}} - {\rm{b}}}}$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{\rm{secnx}}}}{{{\rm{ax}} - {\rm{b}}}}} \right)$

= $\frac{{\left( {{\rm{ax}} - {\rm{b}}} \right)\frac{{{\rm{d}}\left( {{\rm{secnx}}} \right)}}{{{\rm{d}}\left( {{\rm{nx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{nx}}} \right)}}{{{\rm{dx}}}} - {\rm{secnx}}.\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{ax}} - {\rm{b}}} \right){\rm{\: }}}}{{{{\left( {{\rm{ax}} - {\rm{b}}} \right)}^2}}}$

= $\frac{{\left( {{\rm{ax}} - {\rm{b}}} \right).{\rm{secnx}}.{\rm{tannx}}.{\rm{n}} - {\rm{secnx}}.{\rm{a}}}}{{{{\left( {{\rm{ax}} - {\rm{b}}} \right)}^2}}}$ = $\frac{{{\rm{n}}\left( {{\rm{ax}} - {\rm{b}}} \right){\rm{secnx}}.{\rm{tannx}} - {\rm{asecnx}}}}{{{{\left( {{\rm{ax}} - {\rm{b}}} \right)}^2}}}$

 

 

5. Find the differential coefficient of:

(i)$\frac{{1 - 2{{\sin }^2}\frac{{\rm{x}}}{2}}}{{{{\cos }^2}{\rm{x}}}}$ 

Solution:

Or, y = $\frac{{1 - 2{{\sin }^2}\frac{{\rm{x}}}{2}}}{{{{\cos }^2}{\rm{x}}}}$ = $\frac{{{\rm{cosx}}}}{{{{\cos }^2}{\rm{x}}}}$ = $\frac{1}{{{\rm{cosx}}}}$ = secx

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

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

 

(ii)$\frac{{{\rm{sin}}2{\rm{nx}}}}{{{{\cos }^2}{\rm{nx}}}}$

Solution:

Or, y = $\frac{{{\rm{sin}}2{\rm{nx}}}}{{{{\cos }^2}{\rm{nx}}}}$ = $\frac{{2{\rm{sinnx}}.{\rm{cosnx}}}}{{{{\cos }^2}{\rm{nx}}}}$ = 2tan nx

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $2\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan nx) = $\frac{{2\left( {{\rm{d}}\left( {{\rm{tannx}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{nx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{nx}}} \right)}}{{{\rm{dx}}}}$

= 2sec²nx.n. = 2.n.sec²nx

 

(iii) $\frac{{{\rm{sinax}} - {\rm{sinbx}}}}{{{\rm{cosax}} + {\rm{cosbx}}}}$ 

Solution:

Or, y = $\frac{{{\rm{sinax}} - {\rm{sinbx}}}}{{{\rm{cosax}} + {\rm{cosbx}}}}$ = $\frac{{2\cos \frac{{{\rm{ax}} + {\rm{bx}}}}{2}.\sin \frac{{{\rm{ax}} - {\rm{bx}}}}{2}}}{{2\cos \frac{{{\rm{ax}} + {\rm{bx}}}}{2}.\cos \frac{{{\rm{ax}} - {\rm{bx}}}}{2}}}$ = tan $\frac{{{\rm{ax}} - {\rm{bx}}}}{2}$ = tan $\left( {\frac{{{\rm{a}} - {\rm{b}}}}{2}} \right)$x

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

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

= $\frac{{{\rm{d}}\left( {\tan \left( {\frac{{{\rm{a}} - {\rm{b}}}}{2}} \right){\rm{x}}} \right)}}{{{\rm{d}}\left( {\left( {\frac{{{\rm{a}} - {\rm{b}}}}{2}} \right){\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {\left( {\frac{{{\rm{a}} - {\rm{b}}}}{2}} \right){\rm{x}}} \right)}}{{{\rm{dx}}}}$ = sec²$\left( {\frac{{{\rm{a}} - {\rm{b}}}}{2}} \right){\rm{x}}.\left( {\frac{{{\rm{a}} - {\rm{b}}}}{2}} \right).1$

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

 

(iv)$\frac{{1 - {\rm{cosx}}}}{{1 + {\rm{cosx}}}}$

Solution:

Or, y = $\frac{{1 - {\rm{cosx}}}}{{1 + {\rm{cosx}}}}$ = $\frac{{2{{\sin }^2}\frac{{\rm{x}}}{2}}}{{2{{\cos }^2}\frac{{\rm{x}}}{2}}}$ = tan²$\frac{{\rm{x}}}{2}$.

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

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

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

= 2 tan$\frac{{\rm{x}}}{2}$.sec²$\frac{{\rm{x}}}{2}$.$\frac{1}{2}$ = tan$\frac{{\rm{x}}}{2}$.sec²$\frac{{\rm{x}}}{2}$.

 

(v) $\sqrt {\frac{{1 - {\rm{sinx}}}}{{1 + {\rm{sinx}}}}} $

Solution:

Or, y = $\sqrt {\frac{{1 - {\rm{sinx}}}}{{1 + {\rm{sinx}}}}} $ = $\sqrt {\frac{{1 - {\rm{sinx}}}}{{1 + {\rm{sinx}}}}{\rm{*}}\frac{{1 - {\rm{sinx}}}}{{1 - {\rm{sinx}}}}} $ = $\sqrt {\frac{{{{\left( {1 - {\rm{sinx}}} \right)}^2}}}{{{{\cos }^2}{\rm{x}}}}} $ = $\frac{{1 - {\rm{sinx}}}}{{{\rm{cosx}}}}$

= $\frac{{{{\cos }^2}\frac{{\rm{x}}}{2} + {{\sin }^2}\frac{{\rm{x}}}{2} - 2\sin \frac{{\rm{x}}}{2}.\cos \frac{{\rm{x}}}{2}}}{{{{\cos }^2}\frac{{\rm{x}}}{2} - {{\sin }^2}\frac{{\rm{x}}}{2}}}$

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

= $\frac{{1 - \tan \frac{{\rm{x}}}{2}}}{{1 + \tan \frac{{\rm{x}}}{2}}}$    (dividing numerator and denominator by cos $\frac{{\rm{x}}}{2}$).

= $\frac{{\tan \frac{{\rm{\pi }}}{4} - \tan \frac{{\rm{x}}}{2}}}{{1 + \tan \frac{{\rm{\pi }}}{4}.\tan \frac{{\rm{x}}}{2}}}$ = $\tan \left( {\frac{{\rm{\pi }}}{4} - \frac{{\rm{x}}}{2}} \right)$.

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\tan \left( {\frac{{\rm{\pi }}}{4} - \frac{{\rm{x}}}{2}} \right){\rm{\: }}} \right)$

= $\frac{{{\rm{d}}\left( {\tan \left( {\frac{{\rm{\pi }}}{4} - \frac{{\rm{x}}}{2}} \right)} \right)}}{{{\rm{d}}\left( {\frac{{\rm{\pi }}}{4} - \frac{{\rm{x}}}{2}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{\pi }}}{4} - \frac{{\rm{x}}}{2}} \right)}}{{{\rm{dx}}}}$

= sec²$\left( {\frac{{\rm{\pi }}}{4} - \frac{{\rm{x}}}{2}} \right).\left( {\frac{1}{2}} \right)$ = $ - \frac{1}{2}{\sec ^2}\left( {\frac{{\rm{\pi }}}{4} - \frac{{\rm{x}}}{2}} \right)$

 

(vi) $\frac{{{\rm{cos}}2{\rm{x}} + 1}}{{{\rm{sin}}2{\rm{x}}}}$ 

Solution:

Or, y = $\frac{{{\rm{cos}}2{\rm{x}} + 1}}{{{\rm{sin}}2{\rm{x}}}}$ = $\frac{{2{{\cos }^2}{\rm{x}} - 1 + 1}}{{2{\rm{sinx}}.{\rm{cosx}}}}$ = $\frac{{2{{\cos }^2}{\rm{x}}}}{{2{\rm{sinx}}.{\rm{cosx}}}}$ = cotx.

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\cot {\rm{x\: }}} \right)$

= – cosec²x.

 

(vii)$\frac{{{\rm{cos}}2{\rm{x}}}}{{1 - {\rm{sin}}2{\rm{x}}}}$ 

Solution:

Or, y = $\frac{{{\rm{cos}}2{\rm{x}}}}{{1 - {\rm{sin}}2{\rm{x}}}}$ = $\frac{{2{{\cos }^2}{\rm{x}} - {{\sin }^2}{\rm{x}}}}{{{{\cos }^2}{\rm{x}} + {{\sin }^2}{\rm{x}} - 2{\rm{sinx}}.{\rm{cosx}}}}$ = $\frac{{\left( {{\rm{cosx}} + {\rm{sinx}}} \right)\left( {{\rm{cosx}} - {\rm{sinx}}} \right)}}{{{{\left( {{\rm{cosx}} - {\rm{sinx}}} \right)}^2}}}$ = $\frac{{1 + {\rm{tanx}}}}{{1 - {\rm{tanx}}}}$ (dividing numerator and denominator by cosx)

Y = $\frac{{\tan \frac{{\rm{\pi }}}{4} + {\rm{tanx}}}}{{1 - \tan \frac{{\rm{\pi }}}{4}.{\rm{tanx}}}}$ = tan($\frac{{\rm{\pi }}}{4}$ + x)

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{tan}}\left( {\frac{{\rm{\pi }}}{4}{\rm{\: }} + {\rm{\: x}}} \right){\rm{\: }}} \right)$

= $\frac{{{\rm{d}}\left( {{\rm{tan}}\left( {\frac{{\rm{\pi }}}{4}{\rm{\: }} + {\rm{\: x}}} \right)} \right)}}{{{\rm{d}}\left( {\frac{{\rm{\pi }}}{4}{\rm{\: }} + {\rm{\: x}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{\pi }}}{4}{\rm{\: }} + {\rm{\: x}}} \right)}}{{{\rm{dx}}}}$ = sec²$\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right)$.1 = sec²$\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right)$.

 

(viii)$\frac{1}{{{\rm{secx}} - {\rm{tanx}}}}$ 

Solution:

Or, y = $\frac{1}{{{\rm{secx}} - {\rm{tanx}}}}$ = $\frac{1}{{{\rm{secx}} - {\rm{tanx}}}}{\rm{*}}\frac{{{\rm{secx}} + {\rm{tanx}}}}{{{\rm{secx}} + {\rm{tanx}}}}$ = $\frac{{\left( {{\rm{secx}} + {\rm{tanx}}} \right)\^2{\rm{\: }}}}{{{{\sec }^2}{\rm{x}} - {{\tan }^2}{\rm{x}}}}$ = (secx + tanx)²

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\left( {{\rm{secx\: }} + {\rm{\: tanx}}} \right)}^2}} \right)$

= $\frac{{{\rm{d}}\left( {{{\left( {{\rm{secx\: }} + {\rm{\: tanx}}} \right)}^2}{\rm{\: }}} \right)}}{{{\rm{d}}\left( {{\rm{secx\: }} + {\rm{\: tanx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{secx\: }} + {\rm{\: tanx}}} \right)}}{{{\rm{dx}}}}$ = 2(secx + tanx)(secx.tanx + sec²x)

= 2(secx + tanx).secx(tanx + secx) = 2secx(secx + tanx)².

 

(x)$\frac{{1 + {\rm{tanx}}}}{{1 - {\rm{tanx}}}}$

Solution:

Or, y = $\frac{{1 + {\rm{tanx}}}}{{1 - {\rm{tanx}}}}$ = $\frac{{\tan \frac{{\rm{\pi }}}{4} + {\rm{tanx}}}}{{1 - {\rm{tanx}}.\tan \frac{{\rm{\pi }}}{4}}}$ = tan $\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right)$

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

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{tan\: }}\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right){\rm{\: }}} \right)$

= $\frac{{{\rm{d}}\left( {{\rm{tan\: }}\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right){\rm{\: }}} \right)}}{{{\rm{d}}\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right)}}{{{\rm{dx}}}}$ = sec²$\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right)$.1 = sec²$\left( {\frac{{\rm{\pi }}}{4} + {\rm{x}}} \right)$.

 

6. Find the derivative of: 

(i)sin6x.cos4x

Solution:

Let y = sin6x.cos4x = $\frac{1}{2}$(2sin6x.cos4x)

= $\frac{1}{2}${sin(6x + 4x) + sin(6x – 4x)} = $\frac{1}{2}$(sin10x + sin2x)

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{2}\frac{{\rm{d}}}{{{\rm{dx}}}}$(sin 10x + sin2x)

= $\frac{1}{2}\left\{ {\frac{{{\rm{d}}\left( {{\rm{sin}}10{\rm{x}}} \right)}}{{{\rm{d}}\left( {10{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {10{\rm{x}}} \right)}}{{{\rm{dx}}}} + \frac{{{\rm{d}}\left( {{\rm{sin}}2{\rm{x}}} \right)}}{{{\rm{d}}\left( {2{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{x}}} \right)}}{{{\rm{dx}}}}} \right\}$

= $\frac{1}{2}${cos10x.10.1 + cos2x.2.1}

= 5cos10x + cos2x.

 

(ii)sin2mx.sin2nx

Solution:

Let y = sin2mx.sin2nx

= $\frac{1}{2}\left( {2{\rm{sin}}2{\rm{mx}}.{\rm{sin}}2{\rm{nx}}} \right)$

= $\frac{1}{2}\left\{ {\cos \left( {2{\rm{mx}} - 2{\rm{nx}}} \right) - \cos \left( {2{\rm{mx}} + 2{\rm{nx}}} \right)} \right\}$

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{2}\frac{{\rm{d}}}{{{\rm{dx}}}}\left\{ {\cos \left( {2{\rm{mx}} - 2{\rm{nx}}} \right) - \cos \left( {2{\rm{mx}} + 2{\rm{nx}}} \right)} \right\}$

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

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

= (m + n)sin (2m + 2n)x – (m – n)sin(2m – 2n)x

= (m + n)sin2(m + n)x – (m – n)sin2(m – n)x.

 

(iii)cos7x.cos5x

Solution:

Let y = cos7x.cos5x

= $\frac{1}{2}\left( {{\rm{cos}}7{\rm{x}}.{\rm{cos}}5{\rm{x}}} \right)$

= $\frac{1}{2}\left\{ {\cos \left( {7{\rm{x}} + 5{\rm{x}}} \right) + \cos \left( {7{\rm{x}} - 5{\rm{x}}} \right)} \right\}$

= $\frac{1}{2}\left( {{\rm{cos}}12{\rm{x}} + {\rm{cos}}2{\rm{x}}} \right)$

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{2}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{cos}}12{\rm{x}} + {\rm{cos}}2{\rm{x}}} \right)$

= $\frac{1}{2}\left\{ {\frac{{{\rm{d}}\left( {{\rm{cos}}12{\rm{x}}} \right)}}{{{\rm{d}}\left( {12{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {12} \right){\rm{x}}}}{{{\rm{dx}}}} + \frac{{{\rm{d}}\left( {\cos 2{\rm{x}}} \right)}}{{{\rm{d}}\left( {2{\rm{x}}} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{x}}} \right)}}{{{\rm{dx}}}}} \right\}$

= $\frac{1}{2}\left\{ { - {\rm{sin}}12{\rm{x}}.12.1 + \left( { - {\rm{sin}}2{\rm{x}}} \right).2.1} \right\}$

= –6sin12x – sin2x.

 

(iv)sin3x.cos5x

Solution:

Let y = sin3x.cos5x

= $\frac{1}{2}\left( {2{\rm{sin}}3{\rm{x}}.{\rm{cos}}5{\rm{x}}} \right) = \frac{1}{2}\left\{ {{\rm{sin}}\left( {3{\rm{x}} + 5{\rm{x}}} \right) + \sin \left( {3{\rm{x}} - 5{\rm{x}}} \right)} \right\}$

= $\frac{1}{2}\left( {{\rm{sin}}8{\rm{x}} - {\rm{sin}}2{\rm{x}}} \right)$

Differentiating both sides w.r.t. x

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{2}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{sin}}8{\rm{x}} - {\rm{sin}}2{\rm{x}}} \right)$

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

= $\frac{1}{2}\left\{ {{\rm{cos}}8{\rm{x}}.8.1 - {\rm{cos}}2{\rm{x}}.2.1} \right\}$

= 4.cos8x – cos 2x.

 

7. Find $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ , when

(i)x + y = siny

Solution:

Or, x + y = siny

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

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x + y) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sin y)

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x) + $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sin y) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.

Or, 1 + $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = cosy $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, 1 = cosy $\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - \frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, 1 = (cosy – 1)$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

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

 

(ii)x + y = cos(x – y)

Solution:

Or, x + y = cos(x – y)

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

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x + y) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(cos(x – y))

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

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

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

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

Or, 1 + sin(x – y) = {sin(x – y) – 1}$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

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

(iii)x – y = tanxy 

Solution:

Or, x – y = tanxy

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

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x – y) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tanxy)

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)$ – $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left( {{\rm{tanxy}}} \right)}}{{{\rm{d}}\left( {{\rm{xy}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{xy}}} \right)}}{{{\rm{dx}}}}$

Or, 1 – $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = sec² xy $\left\{ {{\rm{x}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)} \right\}$

Or, 1 –$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = xsec²xy $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + ysec²xy.1

Or, 1 – ysec²xy = xsec²xy.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, 1 – ysec²xy = (xsec² xy + 1)$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{1 - {\rm{yse}}{{\rm{c}}^2}{\rm{xy}}}}{{{\rm{xse}}{{\rm{c}}^2}{\rm{xy}} + 1}}$.

 

(iv)x²y = secxy²

Solution:

Or, x²y = secxy²

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

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x²y) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sec xy²)

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

Or, $\frac{{{{\rm{x}}^2}{\rm{dy}}}}{{{\rm{dx}}}}$ + y.2x = sec² xy.tan xy²$\left\{ {{\rm{x}}.\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dx}}}} + {{\rm{y}}^2}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)} \right\}$

Or, x²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}{\rm{\: }}$ + 2xy = sec²xy.tan xy²$\left\{ {{\rm{x}}.\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dx}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{y}}^2}.1} \right\}$.

Or, x²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 2xy = sec xy² tan xy²$\left\{ {{\rm{x}}.2{\rm{y}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{y}}^2}} \right\}$.

Or, x²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 2xy = 2xy.sec xy². tanxy²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y² secxy² tanxy²

Or, 2xy – y²sec xy² tan xy² = 2xy sec xy² tan xy²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ – x²$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, 2xy – y². sec xy². tan xy² = (2xy. secxy² tan xy² – x²) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

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

 

(v)x²y = (ax + by)

Solution:

Or, x²y = (ax + by)

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

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x²y²) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan(ax + by))

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

Or, ${{\rm{x}}^2}\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dx}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {{\rm{y}}^2}.2{\rm{x}}$ = sec² (ax + by) $\left\{ {{\rm{a}}\frac{{{\rm{dx}}}}{{{\rm{dx}}}} + {\rm{b}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right\}$

Or, x²$.$2y.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 2xy² = asec² (ax + b).1 + bsec²(ax + b) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, 2xy² – asec²(ax + by) = bsec²(ax + b) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ – 2x²y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, 2xy² – asec²(ax + by) = {bsec²(ax + by) – 2x²y}$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{2{\rm{x}}{{\rm{y}}^2} - {\rm{a}}.{{\sec }^2}\left( {{\rm{ax}} + {\rm{b}}} \right)}}{{{\rm{bse}}{{\rm{c}}^2}\left( {{\rm{ax}} + {\rm{b}}} \right) - 2{{\rm{x}}^2}{\rm{y}}}}$.

 

(vi) x² + y² = sin xy

Solution:

Or, x² + y² = sin xy

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

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x² + y²) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sin xy)

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

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

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

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

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

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

 

(vii) x²y² = tanxy

Solution:

Or, x²y² = tanxy

Differentiating both sides w.r.t. x

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(x²y²) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan xy)

Or, x²$\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dx}}}}$ + y²$\frac{{{\rm{d}}{{\rm{x}}^2}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{d}}\left( {{\rm{tanxy}}} \right)}}{{{\rm{d}}\left( {{\rm{xy}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{xy}}} \right)}}{{{\rm{dx}}}}$.

Or, x²$\frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y².2x = sec² xy $\left\{ {{\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}}\frac{{{\rm{dx}}}}{{{\rm{dx}}}}} \right\}$

Or, x².2y. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + 2xy² = – xsec²xy $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + ysec² xy.1

Or, 2xy² – ysec² xy = xsec² xy $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ – 2x²y $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.

Or, y(2xy – sec²xy) = –x(2xy – sec²xy)$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.

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

 

(viii) xy = sec (x – y)

Solution:

Or, xy = sec (x – y)

Differentiating both sides w.r.t. x

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(xy) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(sec(x – y))

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

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

Or, x. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y = sec(x – y)tan(x – y) – sec(x – y).tan(x – y) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.

Or, x.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + sec(x – y).tan(x – y) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = sec(x – y).tan(x – y) – y.

Or, {x + sec(x – y).tan(x – y)}$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = sec(x – y).tan(x – y) – y

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{\rm{sex}}\left( {{\rm{x}} - {\rm{y}}} \right)\tan \left( {{\rm{x}} - {\rm{y}}} \right) - {\rm{y}}}}{{{\rm{x}} + \sec \left( {{\rm{x}} - {\rm{y}}} \right).{\rm{tan}}\left( {{\rm{x}} - {\rm{y}}} \right)}}$.

 

(ix) xy = tan(x² + y²)

Solution:

Or, xy = tan(x² + y²)

Differentiating both sides w.r.t. x

Or, $\frac{{\rm{d}}}{{{\rm{dx}}}}$(xy) = $\frac{{\rm{d}}}{{{\rm{dx}}}}$(tan(x² + y²))

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

Or, x $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y.1 = sec² (x² + y²)$\left\{ {\frac{{{\rm{d}}{{\rm{x}}^2}}}{{{\rm{dx}}}} + \frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dx}}}}} \right\}$.

Or, x. $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y = sec²(x² + y²)$\left\{ {2{\rm{x}} + \frac{{{\rm{d}}{{\rm{y}}^2}}}{{{\rm{dy}}}}.\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right\}$

Or, x.$\frac{{{\rm{dy}}}}{{{\rm{dzx}}}}$ + y = sec²(x² + y²) $\left\{ {2{\rm{x}} + 2{\rm{y}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right\}$.

Or, x.$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ + y = 2xsec²(x² + y²) + 2ysec²(x² + y²) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

Or, y – 2xsec²(x² + y²) = 2ysec²(x² + y²) $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ – x$\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$.

Or, y – 2x sec² (x² + y²) = {2ysec²(x² + y²) – x} $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$

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

8. Find $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ , when

(i)x = acos²θ and y = b.sin²θ

Solution:

x = acos²θ

Differentiating both sides w.r.t. θ,

Or, $\frac{{{\rm{dx}}}}{{{\rm{d}}\theta }}$ = a.$\frac{{\rm{d}}}{{{\rm{d}}\theta }}$(cos² θ) = a.$\frac{{{\rm{d}}\left( {{{\cos }^2}\theta } \right)}}{{{\rm{d}}\left( {{\rm{cos}}\theta } \right)}}.\frac{{{\rm{d}}\left( {{\rm{cos}}\theta } \right)}}{{{\rm{d}}\theta }}$

= a.2.cosθ.(–sinθ) = – 2asinθ.cosθ.

y = b.sin²θ

or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = b$\frac{{\rm{d}}}{{{\rm{d}}\theta }}$(sin² θ) = b$\frac{{{\rm{d}}\left( {{{\sin }^2}\theta } \right)}}{{{\rm{d}}\left( {{\rm{sin}}\theta } \right)}}.\frac{{{\rm{d}}\left( {{\rm{sin}}\theta } \right)}}{{{\rm{d}}\theta }}$.

= b.2 sinθ.cosθ = 2b.sinθ.cosθ

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\frac{{{\rm{dy}}}}{{{\rm{d}}\theta }}}}{{\frac{{{\rm{dx}}}}{{{\rm{d}}\theta }}}}$ = $\frac{{2{\rm{bsin}}\theta .{\rm{cos}}\theta }}{{ - 2{\rm{asin}}\theta .{\rm{cos}}\theta }}$ = $ - \frac{{\rm{b}}}{{\rm{a}}}$.

 

 

(ii) x = 2a.sint.cost = asin2t and y = b.cos 2t

Solution:

x = 2a.sint.cost = asin2t

Differentiating both sides w.r.t. t,

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = a.$\frac{{\rm{d}}}{{{\rm{dt}}}}$(sin 2t) = a.$\frac{{{\rm{d}}\left( {{\rm{sin}}2{\rm{t}}} \right)}}{{{\rm{d}}\left( {2{\rm{t}}} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{t}}} \right)}}{{{\rm{dt}}}}$

= a.cos 2t.2.1 = 2a.cos 2t.

y = b.cos 2t
Differentiating both sides w.r.t. θ.

or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = b${\rm{\: }}\frac{{\rm{d}}}{{{\rm{dt}}}}$(cos 2t) = b$\frac{{{\rm{d}}\left( {\cos 2{\rm{t}}} \right)}}{{{\rm{d}}\left( {2{\rm{t}}} \right)}}.\frac{{{\rm{d}}\left( {2{\rm{t}}} \right)}}{{{\rm{dt}}}}$.

= b.(–sin2t).2.1 = –2bsin2t.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}}{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}}$ = $\frac{{ - 2{\rm{bsin}}2{\rm{t}}}}{{2{\rm{acos}}2{\rm{t}}}}$ = $ - \frac{{\rm{b}}}{{\rm{a}}}{\rm{tan}}2{\rm{t}}$.

 

(iii)x = 2a.tanθ and y = a.sec²θ

Solution:

x = 2a.tanθ

Differentiating both sides w.r.t. θ,

Or, $\frac{{{\rm{dx}}}}{{{\rm{d}}\theta }}$ = 2a. $\frac{{\rm{d}}}{{{\rm{d}}\theta }}$(tanθ) = 2asec²θ

y = a.sec²θ
Differentiating both sides w.r.t. θ.

or, $\frac{{{\rm{dy}}}}{{{\rm{d}}\theta }}$ = a.$\frac{{\rm{d}}}{{{\rm{d}}\theta }}$(sec²θ) = a.$\frac{{{\rm{d}}\left( {{{\sec }^2}\theta } \right)}}{{{\rm{d}}\left( {{\rm{sec}}\theta } \right)}}.\frac{{{\rm{d}}\left( {{\rm{sec}}\theta } \right)}}{{{\rm{d}}\theta }}$.

= a.2.secθ.secθ.tanθ = 2a.sec²θ.tanθ

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\frac{{{\rm{dy}}}}{{{\rm{d}}\theta }}}}{{\frac{{{\rm{dx}}}}{{{\rm{d}}\theta }}}}$ = $\frac{{2{\rm{ase}}{{\rm{c}}^2}\theta .{\rm{tan}}\theta }}{{2{\rm{a}}.{{\sec }^2}\theta }}$ = tanθ.

 

(iv) x = tan t and y = sint.cost

Solution:

x = tan t

Differentiating both sides w.r.t. t,

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = $\frac{{\rm{d}}}{{{\rm{dt}}}}$(tant) = sec²t

y = sint.cost
Differentiating both sides w.r.t. θ.

or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = $\frac{{\rm{d}}}{{{\rm{dt}}}}$ (sint.cost) = sint. $\frac{{\rm{d}}}{{{\rm{dt}}}}$(cost) + cost $\frac{{\rm{d}}}{{{\rm{dt}}}}$ (sint)

= sint(–sint) + cost.cost = cos²t – sin²t

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}}{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}}$ = $\frac{{{{\cos }^2}{\rm{t}} - {{\sin }^2}{\rm{t}}}}{{.{{\sec }^2}{\rm{t}}}}$ = cos² t(cos²t – sin²t).

 

(v)x = a(cost + t.sint) and y = a(sint – tcost)

Solution:

x = a(cost + t.sint)

Differentiating both sides w.r.t. t,

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = ${\rm{a}}.\frac{{\rm{d}}}{{{\rm{dt}}}}$(cost + tsint)

= a$\left\{ {\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{\rm{cost}}} \right) + {\rm{t}}.\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{\rm{sint}}} \right) + {\rm{sint}}\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{t}}} \right\}$

= a(–sint + t.cost + sint.1) = at.cost.

y = a(sint – tcost)
Differentiating both sides w.r.t. θ.

or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = ${\rm{a}}.\frac{{\rm{d}}}{{{\rm{dt}}}}$ (sint – tcost)

= ${\rm{a}}\left\{ {\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{\rm{sint}}} \right) - \left( {{\rm{t}}\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{cost}} + {\rm{cost}}\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{t}}} \right)} \right\}$

= a{cost – t(–sint) – cost.1}

= a.cost + at.sint – a.cost = at.sint

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}}{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}}$ = $\frac{{{\rm{at}}.{\rm{sint}}}}{{{\rm{at}}.{\rm{cost}}}}$ = tan.t

 

(vi)x = a(tant – t.sec²t) and y = asec²t

Solution:

x = a(tant – t.sec²t)

Differentiating both sides w.r.t. t,

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}}$ = ${\rm{a}}.\frac{{\rm{d}}}{{{\rm{dt}}}}$(tan.t – t.sec²t)

= ${\rm{a}}\left[ {\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{\rm{tant}}} \right) - \left\{ {{\rm{t}}\frac{{\left( {{\rm{d}}\left( {{{\sec }^2}{\rm{t}}} \right)} \right)}}{{{\rm{d}}\left( {{\rm{sect}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sect}}} \right)}}{{{\rm{dt}}}} + {{\sec }^2}{\rm{t}}\frac{{{\rm{dt}}}}{{{\rm{dt}}}}} \right\}} \right]{\rm{\: }}$

= a(sec³t – (t.2sect.sect.tant + sec²t.1)}

= a(sec²t – 2tsec²tant – sec²t) = –2atsex²t.tant.

y = asec²t
Differentiating both sides w.r.t. t.

or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = ${\rm{a}}.\frac{{\rm{d}}}{{{\rm{dt}}}}$(sec²t) = a$\frac{{{\rm{d}}\left( {{{\sec }^2}{\rm{t}}} \right)}}{{{\rm{d}}\left( {{\rm{sect}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{sect}}} \right)}}{{{\rm{dt}}}}$

= a.2sect.sect.tant = 2asec²t.tant

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}}{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}}$ = $\frac{{2{\rm{a}}.{{\sec }^2}{\rm{t}}.{\rm{tant}}}}{{ - 2.{\rm{at}}.{\rm{se}}{{\rm{c}}^2}.{\rm{t}}.{\rm{tan}}.{\rm{t}}}}$ = $ - \frac{1}{{\rm{t}}}$.

 

(vii)x = a(t + sint) and y = a(1 – cost)

Solution:

x = a(t + sint)

Differentiating both sides w.r.t. t,

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

y = a(1 – cost)

Differentiating both sides w.r.t. t.

or, $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ = ${\rm{a}}.\frac{{\rm{d}}}{{{\rm{dt}}}}$(1 – cost) = a$\left\{ {\frac{{{\rm{d}}\left( 1 \right)}}{{{\rm{dt}}}} - \frac{{\rm{d}}}{{{\rm{dt}}}}\left( {{\rm{cost}}} \right)} \right\}$

= a.{0 – (–sint)} = asint.

Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}}{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}}$ = $\frac{{{\rm{a}}.{\rm{sint}}}}{{{\rm{a}}\left( {1 + {\rm{cost}}} \right)}}$ = $\frac{{{\rm{sint}}}}{{1 + {\rm{cost}}}}$ = $\frac{{2\sin \frac{{\rm{t}}}{2}.\cos \frac{{\rm{t}}}{2}}}{{2{{\cos }^2}\frac{{\rm{t}}}{2}}}$ = tan $\frac{{\rm{t}}}{2}$.

 

9. Differentiate

(i) sin x with respect to cos x.

Solution:

Let y = sinx

Diff. both sides w.r.t. to cosx

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

So, $\frac{{{\rm{dy}}}}{{{\rm{d}}\left( {{\rm{cosx}}} \right)}}$ = – cotx.${\rm{\: \: }}$

 

(ii)tan x with respect to sec x.

Solution:

Let y = tan x.

Diff. both sides w.r.t. to sec x

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

So, $\frac{{{\rm{dy}}}}{{{\rm{d}}\left( {{\rm{cosx}}} \right)}}$ = cosecx.

  

(iii)sec²x with resoect to tan x.

Solution:

Let y = sec²x

Diff. both sides w.r.t. to tanx

Or, $\frac{{{\rm{dy}}}}{{{\rm{d}}\left( {{\rm{tanx}}} \right)}} = \frac{{{\rm{d}}\left( {{{\sec }^2}{\rm{x}}} \right)}}{{{\rm{d}}\left( {{\rm{tanx}}} \right)}} = \frac{{\frac{{{\rm{d}}\left( {{{\sec }^2}{\rm{x}}} \right)}}{{{\rm{dx}}}}}}{{\frac{{{\rm{d}}\left( {{\rm{tanx}}} \right)}}{{{\rm{dx}}}}}}$ = $\frac{{\frac{{{\rm{dse}}{{\rm{c}}^2}{\rm{x}}}}{{{\rm{dsecx}}}}.\frac{{{\rm{dsecx}}}}{{{\rm{dx}}}}}}{{{{\sec }^2}{\rm{x}}}}$ = $\frac{{2{\rm{secx}}.{\rm{secx}}.{\rm{tanx}}}}{{{{\sec }^2}{\rm{x}}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{d}}\left( {{\rm{tanx}}} \right)}}$ = 2tanx.

 

(iv) cosec x with respect to cot x.

Solution:

Let y = cosecx

Diff. both sides w.r.t. to cotx

Or, $\frac{{{\rm{dy}}}}{{{\rm{d}}\left( {{\rm{cotx}}} \right)}} = \frac{{{\rm{d}}\left( {{\rm{cosecx}}} \right)}}{{{\rm{d}}\left( {{\rm{cotx}}} \right)}} = \frac{{\frac{{{\rm{d}}\left( {{\rm{cosecx}}} \right)}}{{{\rm{dx}}}}}}{{{\rm{d}}\left( {{\rm{cotx}}} \right)}}$ = $\frac{{ - {\rm{cosecx}}.{\rm{cotx}}}}{{ - {\rm{cose}}{{\rm{c}}^2}{\rm{x}}}}$ = cos x

So, $\frac{{{\rm{dy}}}}{{{\rm{d}}\left( {{\rm{cotx}}} \right)}}$ = cosx.

 

11. Find the derivative of:

a.cosx°

Solution:

Let y = cosx°

Diff. both sides w.r.t. ‘x’,

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

= – sinx°. $\frac{{{\rm{d}}\left( {\frac{{{\rm{\pi x}}}}{{100}}} \right)}}{{{\rm{dx}}}}$$\left[ {\begin{array}{*{20}{c}}{1\infty  = \frac{{\rm{\pi }}}{{180}}}\\{{\rm{x}}\infty  = \frac{{{\rm{\pi x}}}}{{180}}}\end{array}} \right]$

= –sinx°.$\frac{{\rm{\pi }}}{{180}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $ - \frac{{\rm{\pi }}}{{180}}.{\rm{sinx}}\infty $

 

b.x.sinx°

Solution:

Let y = x.sinx°

Diff. both sides w.r.t. ‘x’,

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

= 1.sinx° + x.$\frac{{{\rm{dsinx}}\infty }}{{{\rm{dx}}\infty }}$.$\frac{{{\rm{dx}}\infty }}{{{\rm{dx}}}}$

= sinx° + x.cosx°. $\frac{{{\rm{d}}\left( {\frac{{{\rm{\pi x}}}}{{180}}} \right)}}{{{\rm{dx}}}}$ = sinx° + xcosx°. $\frac{{\rm{\pi }}}{{180}}$.$\frac{{{\rm{dx}}}}{{{\rm{dx}}}}$

So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = sinx° + $\frac{{\rm{\pi }}}{{180}}$x.cosx°