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  • If <img alt="\left ( \sec x \right )^{y}= \left ( \sin y \right )^{x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%20%28%20%5Csec%20x%20%5Cright%20%29%5E%7By%7D%3D%20%5Cleft%20%28%

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

Option: 1

\frac{\log \sin y+y \cot x}{\log \csc x-x \cot y}


Option: 2

\frac{\log \sin y+y \tan x}{\log \sec x-x \cot y}


Option: 3

\frac{\log \sin y-y \tan x}{\log \sec x-x \cot y}


Option: 4

\frac{\log \cos y+y \cot x}{\log \csc x+x \tan y}


Answers (1)

best_answer

The given term is:

\left ( \sec x \right )^{y}= \left ( \sin y \right )^{x}

Taking logs on both sides,

y \log (\sec x)=x \log (\sin y)

Then differentiating both sides w.r.t. x,
y \cdot \frac{1}{\sec x} \cdot \frac{d}{d x}(\sec x)+\log (\sec x) \frac{d y}{d x}=x \cdot \frac{1}{\sin y} \cdot \frac{d}{d x}(\sin y) \frac{d y}{d x}+\log (\sin y) \cdot 1
y \cdot \frac{1}{\sec x} \cdot(\sec x \tan x)+\log (\sec x) \frac{d y}{d x}=x \cdot \frac{1}{\sin y} \cdot\left(\cos y\right) \frac{d y}{d x}+\log (\sin y)
y \tan x+\log (\sec x) \frac{d y}{d x}=x\cdot \cot y\cdot \frac{d y}{d x}+\log (\sin y)
{\left[\log \sec x-x \cot y\right] \frac{d y}{d x}=\log \sin y-y \tan x}
\frac{dy}{dx}= \frac{\log \sin y-y \tan x}{\log \sec x-x \cot y}

Posted by

Anam Khan

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