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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{\text{e}^\text{t}+\text{e}^{-\text{t}}}{2}\text{ and y}=\frac{\text{e}^\text{t}-\text{e}^\text{-t}}{2}$
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Answer

We have, $ \text{x}=\frac{\text{e}^{\text{t}}+\text{e}^{\text{-t}}}{2}$ and $ \text{y}=\frac{\text{e}^{\text{t}}+\text{e}^{\text{-t}}}{2}$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\bigg[\frac{\text{d}}{\text{dt}}(\text{e}^{\text{t}})+\frac{\text{d}}{\text{dt}}(\text{e}^{\text{-t}})\bigg]$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}\bigg[\frac{\text{d}}{\text{dt}}(\text{e}^{\text{t}})-\frac{\text{d}}{\text{dt}}(\text{e}^{\text{-t}})\bigg]$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=\frac{1}{2}\bigg[\text{e}^{\text{t}}+\text{e}^{\text{-t} \frac{\text{d}}{\text{dt}}}(\text{-t})\bigg]$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}\bigg[\text{e}^{\text{t}}-\text{e}^{\text{-t}}\frac{\text{d}}{\text{dt}}({\text{e}^{\text{-t}}})\bigg]$
$\Rightarrow\frac{1}{2}(\text{e}^{\text{t}}-\text{e}^\text{-t})=\text{y}$ and $\frac{\text{dy}}{\text{dt}}=\frac{1}{2}(\text{e}^\text{t}+\text{e}^{\text{-t}})=\text{x}$
$\therefore\frac{\text{dy}}{\text{dt}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{x}}{\text{y}}$

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