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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\sin^2\text{y}+\cos\text{xy}=\text{k},$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=1,\text{y}=\frac{\pi}{4}$
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Answer

Here, $\text{e}^{\text{x}}+\text{e}^\text{y}=\text{e}^{\text{x}+\text{y}}$
Differentiating with respect to x using chain rule,
$\Rightarrow \frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}\big)+\frac{\text{d}}{\text{dx}}\text{e}^{\text{y}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}+\text{y}}\big)$
$\Rightarrow \text{e}^\text{x}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dy}}(\text{a}+\text{y})$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\Rightarrow\text{e}^{\text{x}}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big(\text{e}^{\text{y}}-\text{e}^{\text{x}+\text{y}}\big)=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{e}^\text{x}\times\text{e}^\text{y}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^\text{x}\times\text{e}^\text{y}}\Big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\big(\text{e}^\text{y}-1\big)}{\text{e}^\text{y}\big({1-\text{e}}^\text{x}\big)}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=-\frac{\text{e}^\text{x}\big(\text{e}^\text{y}-1\big)}{\text{e}^\text{y}\big({\text{e}^\text{x}-1}\big)}$

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