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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If $\text{x}=\text{e}^{\cos2\text{t}}$ and $\text{y}=\text{e}^{\sin2\text{t}},$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$
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Answer

We have, $\text{x}=(\text{e}^{\cos2\text{t}})$ and $\text{y}=\text{e}^{\sin2\text{t}}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\big(\text{e}^{\cos2\text{t}}\big)$ and $\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dt}}\big(\text{e}^{\sin2\text{t}}\big)$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})$ and $\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}\frac{\text{d}}{\text{dt}}(\cos2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\text{e}^{\cos2\text{t}}(-\sin2\text{t})\frac{\text{d}}{\text{dt}}(2\text{t})$ and $\frac{\text{dy}}{\text{dt}}=\text{e}^{\sin2\text{t}}(\cos2\text{t})\frac{\text{d}}{\text{dt}}(2\text{t})$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=-2\sin 2\text{t}\text{e}^{\cos2\text{t}}$ and $\frac{\text{dy}}{\text{dt}}=2\cos 2\text{t}\text{e}^{\sin2\text{t}}$
$\because\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\cos2\text{te}^{\sin2\text{t}}}{-2\sin2\text{te}^{\cos2\text{t}}}$
$\begin{bmatrix} \because\text{x}=\text{e}^{\cos2\text{t}}\Rightarrow\log\text{x}=\cos2\text{t} \\ \text{y}=\text{e}^{\sin2\text{t}}\Rightarrow\log\text{y}=\sin2\text{t} \end{bmatrix}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}\log\text{x}}{\text{x}\log\text{y}}$

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