#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.5 question 41 maths textbook solution

Answer: $\frac{d y}{d x}=\frac{\log (\operatorname{cosy})-y \cot x}{\log (\sin x)+x \tan y}$

Hint:  To solve this equation we will convert log function

Given:  $(\sin x)^{y}=(\cos y)^{x}$

Solution:

\begin{aligned} &\log (\sin x)^{y}=\log (\cos y)^{x} \\\\ &y\log (\sin x)=x \log (\cos y) \end{aligned}

\begin{aligned} &y \frac{d}{d x} \log (\sin x)+\log \sin x \frac{d y}{d x}=x \frac{d}{d x} \log \cos y+\log \cos y \frac{d x}{d x} \\\\ &y \frac{1}{\sin x} \cos x+\log \sin x \frac{d y}{d x}=x \frac{1}{\cos y}(-\sin x) \frac{d y}{d x}+\log \cos y \cdot 1 \end{aligned}

\begin{aligned} &y \cot x+\log \sin x \frac{d y}{d x}=-x \tan \frac{d y}{d x}+\log \cos y \\\\ &\frac{d x}{d x} \log (\sin x)+y \frac{\cos x}{\sin x}=\log (\cos y)+x \frac{-\sin y}{\cos y} \end{aligned}

\begin{aligned} &\frac{d x}{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} \end{aligned}

Hence it is proved