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 Q14  Find dy/dx of the functions given in Exercises 12 to 15. ( \cos x )^y = ( \cos y )^x

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Given function is
f(x)\Rightarrow (\cos x) ^ y = (\cos y) ^ x
Now, take log on both the sides
y\log \cos x = x \log \cos y
Now, differentiate w.r.t  x
\frac{dy}{dx}(\log \cos x)-y\tan x = \log \cos y-x\tan y\frac{dy}{dx}
By taking similar terms on the same side
We get, 
(\frac{dy}{dx}(\log \cos x)-y\tan x) = (\log \cos y-x\tan y\frac{dy}{dx})\\ \frac{dy}{dx} \left (\log \cos x+(\cos y)^x.x\tan y) \right )= \left ( \log \cos y+(\cos x)^y.y\tan x \right )\\ \frac{dy}{dx}= \frac{\left (\log \cos y+y\tan x \right )}{\left ( \log \cos x+x\tan y) \right )} = \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}

Therefore, the answer is  \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x} 

Posted by

Gautam harsolia

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