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Question:

If $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos t 2 t}}$, find $\frac{d y}{d x}$

Solution:

We have, $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}$ and $y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$

$\Rightarrow \frac{d x}{d t}=\frac{d}{d t}\left[\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}\right]$

$\Rightarrow \frac{d x}{d t}=\frac{\sqrt{\cos 2 t} \frac{d}{d t}\left(\sin ^{3} t\right)-\sin ^{3} t \frac{d}{d t} \sqrt{\cos 2 t}}{\cos 2 t}$            [Using quotient rule]

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

$\Rightarrow \quad \frac{d x}{d t}=\frac{3 \sqrt{\cos 2 t}\left(\sin ^{2} t \cos t\right)-\frac{\sin ^{3} t}{2 \sqrt{\cos 2 t}}(-2 \sin 2 t)}{\cos 2 t}$

$\Rightarrow \quad \frac{d x}{d t}=\frac{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}$

Now, $\frac{d y}{d t}=\frac{d}{d t}\left[\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}\right]$

$\Rightarrow \quad \frac{d y}{d t}=\frac{\sqrt{\cos 2 t} \frac{d}{d t}\left(\cos ^{3} t\right)-\cos ^{3} t \frac{d}{d t} \sqrt{\cos 2 t}}{\cos 2 t}$             [Using quotient rule]

$\Rightarrow \quad \frac{d y}{d t}=\frac{\sqrt{\cos 2 t}\left(3 \cos ^{2} t\right) \frac{d}{d t}(\cos t)-\cos ^{3} t \times \frac{1}{2 \sqrt{\cos 2 t}} \frac{d}{d t}(\cos 2 t)}{\cos 2 t}$

$\Rightarrow \quad \frac{d y}{d t}=\frac{3 \sqrt{\cos 2 t} \cos ^{2} t(-\sin t)-\frac{\cos ^{3} t}{2 \sqrt{\cos 2 t}}(-2 \sin 2 t)}{\cos 2 t}$

$\Rightarrow \quad \frac{d y}{d t}=\frac{-3 \cos 2 t \cos ^{2} t \sin t+\cos ^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-3 \cos 2 t \cos ^{2} t \sin t+\cos ^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}} \times \frac{\cos 2 t \sqrt{\cos 2 t}}{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t \sin 2 t}$

$\Rightarrow \quad \frac{d y}{d x}=\frac{-3 \cos 2 t \cos ^{2} t \sin t+\cos ^{3} t \sin 2 t}{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t \sin 2 t}$

$\Rightarrow \quad \frac{d y}{d x}=\frac{\sin t \cos t\left[-3 \cos 2 t \cos t+2 \cos ^{3} t\right]}{\sin t \cos t\left[3 \cos 2 t \sin t+2 \sin ^{3} t\right]}$

$\Rightarrow \quad \frac{d y}{d x}=\frac{\left[-3\left(2 \cos ^{2} t-1\right) \cos t+2 \cos ^{3} t\right]}{\left[3\left(1-2 \sin ^{2} t\right) \sin t+2 \sin ^{3} t\right]}$            $\left[\begin{array}{l}\cos 2 t=2 \cos ^{2} t-1 \\ \cos 2 t=1-2 \sin ^{2} t\end{array}\right]$

$\Rightarrow \quad \frac{d y}{d x}=\frac{-4 \cos ^{3} t+3 \cos t}{3 \sin t-4 \sin ^{3} t}$

$\Rightarrow \quad \frac{d y}{d x}=\frac{-\cos 3 t}{\sin 3 t}$               $\left[\begin{array}{c}\cos 3 t=4 \cos ^{3} t-3 \cos t \\ \sin 3 t=3 \sin t-4 \sin ^{3} t\end{array}\right]$

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

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