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5. Differentiate the following w.r.t. x: 

\log (\cos e ^x )

Answers (1)

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Given function is
f(x)=\log (\cos e ^x )
Let's take g(x ) = \cos e^{x}
Now, our function reduces to 
f(x) = \log(g(x))
Now,
f^{'}(x) = g^{'}(x).\frac{1}{g(x)}                   -(i)
And
g(x)=\cos e^{x}\\\Rightarrow g^{'}(x) = \frac{d(\cos e^{x})}{dx}.\frac{d(e^{x})}{dx}= (-\sin e^x).e^{x} = -e^x.\sin e^x 
Put this value in our equation (i)
f^{'}(x) =-e^x.\sin e^x.\frac{1}{\cos e^x} = -e^x.\tan e^x \ \ \ \ \ (\because \frac{\sin x}{\cos x}=\tan x)
Therefore, the answer is -e^x.\tan e^x,\ \ \ e^x\neq (2n+1)\frac{\pi}{2},\ \ n\in N

Posted by

Gautam harsolia

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