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

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

Answers (1)

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

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions