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  • Differentiate the following w.r.t. x: cos log x + e ^ x

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

\cos ( log x + e ^x ) , x > 0

Answers (1)

best_answer

Given function is
f(x)=\cos ( log x + e ^x )
Lets take g(x) = ( log x + e ^x )
Then , our function reduces to
f(x) = \cos (g(x))
Now, differentiation w.r.t. x is
f^{'}(x) = g^{'}(x)\(-\sin) (g(x))                            -(i)
And
g(x) = ( log x + e ^x )
g^{'}(x)= \frac{d(\log x)}{dx}+\frac{d(e^x)}{dx} = \frac{1}{x}+e^x
Put this value in our equation (i)
f^{'}(x) = -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x)
Therefore, the answer is -\left ( \frac{1}{x}+e^x \right )\sin (\log x+e^x), x>0

Posted by

Gautam harsolia

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