Differentiation Question 140
Question: If $ f(x)={{( \frac{{{\sin }^{m}}x}{{{\sin }^{n}}x} )}^{m+n}}.{{( \frac{{{\sin }^{n}}x}{{{\sin }^{p}}x} )}^{n+p}}.{{( \frac{{{\sin }^{p}}x}{{{\sin }^{m}}x} )}^{p+m}} $ Then $ f’(x) $ is equal to
Options:
A) 0
B) 1
C) $ {{\cos }^{m+n+px}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] we have, $ f(x)={{(si{n^{m-n}}x)}^{m+n}}.{{(si{n^{n-p}}x)}^{n+p}}.{{(si{n^{p-m}}x)}^{p+m}} $
$ ={{\sin }^{m^{2}-n^{2}}}x.{{\sin }^{n^{2}-p^{2}}}x.{{\sin }^{p^{2}-m^{2}}}x $
$ ={{(sinx)}^{m^{2}-n^{2}+n^{2}-p^{2}+p^{2}-m^{2}}}={{(sinx)}^{0}}=1. $
$ \therefore f’(x)=0. $
BETA
