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If \left ( \csc x \right )^{y}= \left ( \cot y \right )^{x} find \frac{dy}{dx}.

Option: 1

\frac{\log \cot y-y \tan x}{\log \sec x+x \frac{\csc ^2 y}{\cot y}}


Option: 2

\frac{\log \cot y-y \tan x}{\log \cos x-x \frac{\csc ^2 y}{\cot y}}


Option: 3

\frac{\log \cot y-y \frac{1}{\sin x \cos x}}{\log \tan x+x \frac{\csc ^2 y}{\cot y}}


Option: 4

\frac{\log \cot y+y \cot x}{\log \csc x+x \frac{\csc ^2 y}{\cot y}}


Answers (1)

best_answer

The given term is:

\left ( \csc x \right )^{y}= \left ( \cot y \right )^{x}

Taking logs on both sides,

y \log (\csc x)=x \log (\cot y)

Then differentiating both sides w.r.t. x,

y \cdot \frac{1}{\csc x} \cdot \frac{d}{d x}(\csc x)+\log (\csc x) \frac{d y}{d x}=x \cdot \frac{1}{\cot y} \cdot \frac{d}{d x}(\cot y) \frac{d y}{d x}+\log (\cot y) \cdot 1
y \cdot \frac{1}{\csc x} \cdot(-\csc x \cot x)+\log (\csc x) \frac{d y}{d x}=x \cdot \frac{1}{\cot y} \cdot\left(-\csc ^2 y\right) \frac{d y}{d x}+\log (\cot y)
-y \cot x+\log (\csc x) \frac{d y}{d x}=-x \frac{\csc ^2 y}{\cot y}\cdot \frac{d y}{d x}+\log (\cot y)
{\left[\log \csc x+x \frac{\csc { }^2 y}{\cot y}\right] \frac{d y}{d x}=\log \cot y-y \cot x}
\frac{dy}{dx}= \frac{\log \cot y+y \cot x}{\log \csc x+x \frac{\csc ^2 y}{\cot y}}

 

Posted by

shivangi.shekhar

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