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Question:

If $(\sin x)^{y}=(\cos y)^{x}$, prove that $\frac{d y}{d x}=\frac{\log \cos y-y \cot x}{\log \sin x+x \tan y}$

Solution:

Here,

$(\sin x)^{y}=(\cos y)^{x}$

Taking log on both sides,

$\log (\sin x)^{y}=\log (\cos y)^{x}$

$y \log (\sin x)=x \log (\cos y)\left[U \operatorname{sing} \log a^{b}=b \log a\right]$

Differentiating it with respect to $x$ using product rule and chain rule,

$\frac{d}{d x}[y \log \sin x]=\frac{d}{d x}[x \log \cos y]$

$y \frac{d}{d x}(\log \sin x)+\log \sin x \frac{d y}{d x}=x \frac{d y}{d x} \log \cos y+\log \cos y \frac{d}{d x}(x)$

$y\left(\frac{1}{\sin x}\right) \frac{d}{d x}(\sin x)+\log \sin x \frac{d y}{d x}=\frac{x}{\cos y} \frac{d}{d x}(\cos y)+\log \cos y(1)$

$\frac{y}{\sin x}(\cos x)+\log \sin x \frac{d y}{d x}=\frac{x}{\cos y}(-\sin y) \frac{d y}{d x}+l \log \cos y$

$y \cot x+\log \sin x \frac{d y}{d x}=-x \tan y \frac{d y}{d x}+\log \cos y$

$\log \sin x \frac{d y}{d x}+x \tan y \frac{d y}{d x}=\log \cos y-y \cot x$

$\frac{d y}{d x}(\log \sin x+x \tan y)=\log \cos y-y \cot x$

$\frac{d y}{d x}=\frac{\log \cos y-y \cot x}{\log \sin x+x \tan y}$

Hence Proved.

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