# Find of function.

Question:

Find $\frac{d y}{d x}$ of function.

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

Solution:

The given function is $(\cos x)^{y}=(\cos y)^{x}$

Taking logarithm on both the sides, we obtain

$y \log \cos x=x \log \cos y$

Differentiating both sides, we obtain

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

$\Rightarrow \log \cos x \frac{d y}{d x}+y \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)=\log \cos y \cdot 1+x \cdot \frac{1}{\cos y} \cdot \frac{d}{d x}(\cos y)$

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

$\Rightarrow \log \cos x \frac{d y}{d x}-y \tan x=\log \cos y-x \tan y \frac{d y}{d x}$

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

$\therefore \frac{d y}{d x}=\frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}$