Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 19

$\frac{d y}{d x}=-\cot 3 t$

Hint:

Use chain rule, product rule and  $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given:

\begin{aligned} &x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}} \\ &y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}} \end{aligned}

Solution:

$x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}} \\$

$x=\sin ^{3} t(\cos 2 t)^{\frac{-1}{2}}\\$                                                                 $\left[\because \frac{1}{x^{n}}=(x)^{-n}\right]$

\begin{aligned} & &\frac{d x}{d t}=(\cos 2 t)^{\frac{-1}{2}} \frac{d \sin ^{3} t}{d t}+\sin ^{3} t \times \frac{d(\cos 2 t)^{\frac{-1}{2}}}{d t} \end{aligned}                                                                  [Using product rule]

$=\left[(\cos 2 t)^{\frac{-1}{2}} \times \frac{d \sin ^{3} t}{d \sin t} \times \frac{d \sin t}{d t}\right]+\left[\sin ^{3} t \times \frac{d(\cos 2 t)^{\frac{-1}{2}}}{d \cos 2 t} \times \frac{d \cos 2 t}{d t}\right]$                                                                 [Using chain rule]

$\frac{d x}{d t}=\left[\frac{1}{\sqrt{\cos 2 t}} \times 3 \sin ^{2} t\right]+\left[\sin ^{3} t \times\left(\frac{-1}{2}(\cos 2 t)^{\frac{-1}{2}-1}\right) \times(-2 \sin 2 t)\right]$

$\frac{d x}{d t}=\frac{3 \sin ^{2} t \cos t}{\sqrt{\cos 2 t}}+\frac{\sin ^{3} t \cdot \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \\$

\begin{aligned} & &\frac{d x}{d t}=\frac{3 \sin ^{2} t \cos t}{\sqrt{\cos 2 t}}+\frac{\sin ^{3} t \cdot \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}                                                                                               (1)

$=\frac{3 \sin ^{2} t \cos t \cdot \cos 2 t+\sin ^{3} t \cdot \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \\$

\begin{aligned} &=\frac{3 \sin ^{2} t \cos t\left(1-2 \sin ^{2} t\right)+\sin ^{3} t(2 \sin t \cdot \cos t)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}                                                  $\left[\begin{array}{l} \because \cos 2 \theta=1-2 \sin ^{2} \theta \\ \sin 2 \theta=2 \sin \theta \cos \theta \end{array}\right]$

$=\frac{3 \sin ^{2} t \cos t-4 \sin ^{4} t \cos t+2 \sin ^{4} t \cdot \cos t}{(\cos 2 t)^{\frac{3}{2}}} \\$

$\frac{d x}{d t}=\frac{3 \sin ^{2} t \cos t-4 \sin ^{4} t \cos t}{(\cos 2 t)^{\frac{3}{2}}} \\$

\begin{aligned} & &=\frac{\sin t \cos t\left(3 \sin t-4 \sin ^{3} t\right)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}

$=\frac{\sin t \cos t \cdot \sin 3 t}{(\cos 2 t)^{\frac{3}{2}}}$                                         $\left[\because \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta\right]$

Multiply and divide by 2

$\frac{d x}{d t}=\frac{2 \sin t \cos t \sin 3 t}{2(\cos 2 t)^{\frac{3}{2}}} \\$

\begin{aligned} & &\frac{d x}{d t}=\frac{\sin 2 t \cdot \sin 3 t}{2(\cos 2 t)^{\frac{3}{2}}} \end{aligned}

$\frac{d x}{d t}=\frac{\sin 2 t \cdot \sin 3 t}{2(\cos 2 t)^{\frac{3}{2}}} \\$                                                                                                                    (2)

\begin{aligned} & &y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}} \end{aligned}

$y=\cos ^{3} t(\cos 2 t)^{\frac{-1}{2}}$                                     $\left[\because \frac{1}{x^{n}}=(x)^{-n}\right]$

\begin{aligned} & \\ &y=\cos ^{3} t(\cos 2 t)^{\frac{-1}{2}} \end{aligned}

$\frac{d y}{d x}=\cos ^{3} t \times \frac{d(\cos 2 t)^{\frac{-1}{2}}}{d t}+(\cos 2 t)^{\frac{-1}{2}} \frac{d \cos ^{3} t}{d t}$                                                                  [Using product rule]

$=\left[\cos ^{3} t \times \frac{d(\cos 2 t)^{\frac{-1}{2}}}{d \cos 2 t} \times \frac{d \cos 2 t}{d(2 t)} \times \frac{d(2 t)}{d t}\right]+\left[(\cos 2 t)^{\frac{-1}{2}} \times \frac{\cos ^{3} t}{d \cos t} \times \frac{d \cos t}{d t}\right]$                                                                                                                                                          [Using chain rule]

$=\left[\cos ^{3} t \times\left(\frac{-1}{2}(\cos 2 t)^{\frac{-3}{2}}\right) \times(-\sin 2 t) \times 2\right]+\left[\frac{1}{\sqrt{\cos 2 t}} \times 3 \cos ^{2} t \times(-\sin t)\right]$

$=\frac{-3 \cos ^{2} t \sin t}{\sqrt{\cos 2 t}}+\frac{\cos ^{3} t \times \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \\$

\begin{aligned} & &=\frac{-3 \cos ^{2} t \sin t(\cos 2 t)+\cos ^{3} t \times \sin 2 t}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}

$=\frac{-3 \cos ^{2} t \cdot \sin t\left(2 \cos ^{2} t-1\right)+\cos ^{3} t \times 2 \sin t \times \cos t}{(\cos 2 t)^{\frac{3}{2}}}$                                                   $\left[\begin{array}{l} \because \cos 2 \theta=2 \cos ^{2} \theta-1 \\ \sin 2 \theta=2 \sin \theta \cos \theta \end{array}\right]$

$\frac{d y}{d t}=\frac{-3 \cos ^{2} t\left(2 \sin t \cos ^{2} t\right)+3 \cos ^{2} t \sin t+2 \cos ^{4} t \sin t}{(\cos 2 t)^{\frac{3}{2}}}$

$=\frac{\left(-6 \cos ^{4} t \sin t+2 \cos ^{4} t \sin t\right)+3 \cos ^{2} t \sin t}{(\cos 2 t)^{\frac{3}{2}}} \\$

\begin{aligned} & &=\frac{-4 \cos ^{4} t \sin t+3 \cos ^{2} t \sin t}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}

$=\frac{-\sin t \cos t\left(4 \cos ^{3} t-3 \cos t\right)}{(\cos 2 t)^{\frac{3}{2}}} \\$

\begin{aligned} & &=\frac{-\sin t \cos t(\cos 3 t)}{(\cos 2 t)^{\frac{3}{2}}} \end{aligned}                                                   $\left[\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\right]$

$=\frac{-2 \sin (-\cos t \cdot \cos 3 t)}{2(\cos 2 t)^{\frac{3}{2}}} \\$

\begin{aligned} & &\frac{d y}{d t}=\frac{-\sin 2 t \cdot \cos 3 t}{2(\cos 2 t)^{\frac{3}{2}}} \end{aligned}                                                                                                                                                    (3)

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Put the values of  $\frac{d y}{d t} \text { and } \frac{d x}{d t}$   from equation (2) and (1) respectively

$\frac{d y}{d x}=\frac{\frac{(-\sin 2 t \times \cos 3 t)}{2(\cos 2 t)^{\frac{3}{2}}}}{\frac{(\sin 2 t \times \sin 3 t)}{2(\cos 2 t)^{\frac{3}{2}}}}$

$=\frac{(-\sin 2 t \times \cos 3 t) \times 2(\cos 2 t)^{\frac{3}{2}}}{(\sin 2 t \times \cos 3 t) \times(2 \cos 2 t)^{\frac{3}{2}}} \\$

$=\frac{-\cos 3 t}{\sin 3 t} \\$

\begin{aligned} & &\frac{d y}{d x}=-\cot 3 t \end{aligned}