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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{x}=\frac{\sin^3\text{t}}{\sqrt{\cos^2\text{t}}},\text{y}=\frac{\cos^3\text{t}}{\sqrt{\cos2\text{t}}},$ find $\frac{\text{dy}}{\text{dx}}$
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Answer

We have, $\text{x}=\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}$ and $\text{y}=\frac{\cos^{3}\text{t}}{\sqrt{\cos2\text{t}}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\sin^{3}\text{t}}{\sqrt{\cos2\text{t}}}\Big] $
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\sin^{3})-\sin^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\sin^{3}\text{t})\frac{\text{d}}{\text{dt}}(\sin\text{t})-\sin^{3}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}(\sin^{2}\text{t}\cos\text{t})-\frac{\sin^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{3\cos2\text{t}\sin^{\text{2}}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
Now, $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\Big[\frac{\cos^{3}}{\sqrt{\cos2\text{t}}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}} \frac{\text{d}}{\text{dt}}(\cos^{3}\text{t})-\cos^{3}\text{t}\frac{\text{d}}{\text{dt}}\sqrt{\cos2\text{t}}}{\cos2\text{t}}$
[Using quotient rule]
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sqrt{\cos2\text{t}}(3\cos^{2}\text{t})\frac{\text{d}}{\text{dt}}(\cos\text{t})-\cos^{3}\text{t}\times\frac{1}{2\sqrt{\cos2\text{t}}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{3\sqrt{\cos2\text{t}}\cos^{2}\text{t}-(\sin\text{t})-\frac{\cos^{3}\text{t}}{2\sqrt{\cos2\text{t}}}(-2\sin2\text{t})}{\cos2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-3\cos2\text{t}\cos^{2}\text{t}\sin\text{t}+\cos^{3}\text{t}\sin2\text{t}}{\cos2\text{t}\sqrt{\cos2\text{t}}}\times\frac{\cos2\text{t}\sqrt{\cos2\text{t}}}{3\cos2\text{t}\sin^{2}\text{t}\cos\text{t}+\sin^{3}\text{t}\sin2\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{\sin\text{t}\cos{\text{t}}[-3\cos2\text{t}\cos\text{t}+2\cos^{3}\text{t}]}{\sin\text{t}\cos\text{t}[3\cos2\text{t}\sin{\text{t}}+2\sin^{3}\text{t}]}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{[-3(2\cos^{2}\text{t}-1)\cos\text{t}+2\cos^{3}\text{t}]}{[3(1-2\sin^{2}\text{t})\sin\text{t}+2\sin^{3}\text{t}]}$
$\begin{bmatrix} \cos2\text{t}=2\cos^2\text{t}-1 \\ \cos2\text{t}=1-2\sin^2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-4\cos^{3}\text{t}+3\cos\text{t}}{3\sin\text{t}-4\sin^{3}\text{t}}$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=\frac{-\cos3\text{t}}{\sin3\text{t}}$
$\begin{bmatrix} \cos3\text{t}=4\cos^3\text{t}-3\cos\text{t} \\ \sin3\text{t}=3\sin^2\text{t}-4\sin^3\text{t} \end{bmatrix}$
$\therefore\frac{\text{dy}}{\text{dx}}=-\cos3\text{t}$

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