If (cos x) y = (sin y) x , then dydx=

(cos⁡x)y=(sin⁡y)x ⇒ylog⁡cos⁡x=xlog⁡sin⁡y y1cos⁡x(−sin⁡x)+log⁡cos⁡xdydx=x1sin⁡y(cos⁡y)dydx+log⁡sin⁡y(1) −ytan⁡x+log⁡cos⁡xdydx=xcot⁡ydydx+log⁡sin⁡y(1)log⁡cos⁡xdydx−xcot⁡ydydx=log⁡sin⁡y+ytan⁡xdydx(log⁡cos⁡x−xcot⁡y)=log⁡sin⁡y+ytan⁡x dydx=log⁡sin⁡y+ytan⁡xlog⁡cos⁡x−xcot⁡y

If (cos x) y = (sin y) x , then dydx=

(cos⁡x)y=(sin⁡y)x ⇒ylog⁡cos⁡x=xlog⁡sin⁡y y1cos⁡x(−sin⁡x)+log⁡cos⁡xdydx=x1sin⁡y(cos⁡y)dydx+log⁡sin⁡y(1) −ytan⁡x+log⁡cos⁡xdydx=xcot⁡ydydx+log⁡sin⁡y(1)log⁡cos⁡xdydx−xcot⁡ydydx=log⁡sin⁡y+ytan⁡xdydx(log⁡cos⁡x−xcot⁡y)=log⁡sin⁡y+ytan⁡x dydx=log⁡sin⁡y+ytan⁡xlog⁡cos⁡x−xcot⁡y

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If (cos x)^y=(sin y) ^x , then $\frac{d y}{d x} =$

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$\frac{\log \sin y - y \tan x}{\log \cos x + c o t y}$

$\frac{\log \cos x}{\log \sin y}$

$\frac{\log \sin y}{\log \cos x}$

$\frac{\log \sin y + y \tan x}{\log \cos x - x c o t y}$

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Detailed Solution

$(\cos x)^{y} = (\sin y)^{x} \Rightarrow y \log \cos x = x \log \sin y y \frac{1}{\cos x} (- \sin x) + \log \cos x \frac{d y}{d x} = x \frac{1}{\sin y} (\cos y) \frac{d y}{d x} + \log \sin y (1)$
$\begin{array}{l} - y \tan x + \log \cos x \frac{d y}{d x} = x \cot y \frac{d y}{d x} + \log \sin y (1) \\ \log \cos x \frac{d y}{d x} - x \cot y \frac{d y}{d x} = \log \sin y + y \tan x \\ \frac{d y}{d x} (\log \cos x - x \cot y) = \log \sin y + y \tan x \end{array}$
$\frac{d y}{d x} = \frac{\log \sin y + y \tan x}{\log \cos x - x \cot y}$