Differentiation Question 105
Question: The first derivative of the function $ (\sin 2x\cos 2x\cos 3x+{\log_2}{2^{x+3}}) $ with respect to x at $ x=\pi $ is
[MP PET 1998]
Options:
A) 2
B) -1
C) $ -2+{2^{\pi }}{\log _{e}}2 $
D) $ -2+{\log _{e}}2 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=\sin 2x.\cos 2x.\cos 3x+{\log_2}{2^{x+3}} $
$ f(x)=\frac{1}{2}\sin 4x\cos 3x+(x+3){\log_2}2 $
$ f(x)=\frac{1}{4}[\sin 7x+\sin x]+x+3 $
Differentiate w.r.t. x, $ f’(x)=\frac{1}{4}[7\cos 7x+\cos x]+1 $
$ f’(x)=\frac{7}{4}\cos 7x+\frac{1}{4}\cos x+1 $ . Hence $ f’(\pi )=-2+1=-1 $ .
BETA
