If (\cos x)^y=(\sin y)^x, find \frac{\mathrm{d} y}{\mathrm{d} x}

 

 

 

 
 
 
 
 

Answers (1)

(\cos x)^y=(\sin y)^x

on taking log on both sides, we get

y\log (\cos x)=x log (\sin y)\: \: \: \: -(i)

On differentiating equation (i) w.r.t x

\Rightarrow y \times \frac{\mathrm{d} }{\mathrm{d} x}(\log \cos x)+\log \cos x\frac{\mathrm{d} y}{\mathrm{d} x}=\log \sin y \frac{\mathrm{d} x}{\mathrm{d} x}+x\frac{\mathrm{d} }{\mathrm{d} x}(\log \sin y)

\Rightarrow y \times \frac{1}{\cos x}(-\sin x)+\log \cos x\frac{\mathrm{d} y}{\mathrm{d} x}=\log \sin y + x \cdot \frac{\cos y}{\sin y }\frac{\mathrm{d} y}{\mathrm{d} x}

\Rightarrow -y \ tanx + \frac{\mathrm{d} y}{\mathrm{d} x}\log \cos x = \log \sin y + x \cot y \frac{\mathrm{d} y}{\mathrm{d} x}

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

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

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