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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\sin^{2}\text{y}+\cos\text{xy}=\pi$
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Answer

The given relationship is $\sin^{2}\text{y}+\cos\text{xy}=\pi$
differenting this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\sin^{2}\text{y}+\cos\text{xy})=\frac{\text{d}}{\text{dx}}(\pi)$
$\Rightarrow\frac{\text{d}}{\text{dx}}(\sin^{2}\text{y})+\frac{\text{d}}{\text{dx}}(\cos\text{xy})=0\ ...(\text{i})$
Using chain rule, we obtain
$\frac{\text{d}}{\text{dx}}(\sin^{2}\text{y})= 2\sin\text{y}\frac{\text{d}}{\text{dx}}(\sin\text{y})=2\sin\text{y}\cos\text{y}\frac{\text{dy}}{\text{dx}} ...\text{(ii)}$
$\frac{\text{d}}{\text{dx}}(\cos\text{xy})=-\sin\text{xy}\frac{\text{d}}{\text{dx}}(\text{xy})=-\sin\text{xy}\Big[\text{y}\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{dy}}{\text{dx}}\Big]$
$=-\sin\text{xy}\Big[\text{y}.1+\text{x}\frac{\text{dy}}{\text{dx}}\Big]= -\text{y}\sin\text{xy}-\text{x}\sin\text{xy}\frac{\text{dy}}{\text{dy}} ...(\text{iii})$
From (1), (2) and (3), we obtain
$2\sin\text{y}\cos\text{y}\frac{\text{dy}}{\text{dx}}-\text{y}\sin\text{xy}-\text{x}\sin\text{xy}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow(2\sin\text{y}\cos\text{y}-\text{x}\sin\text{xy})\frac{\text{dy}}{\text{dx}}=\text{y}\sin\text{xy}$

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