Differentiation Question 303
Question: $ \frac{d}{dx}[ {{\tan }^{-1}}( \frac{a-x}{1+ax} ) ]= $
[Karnataka CET 2001; Pb. CET 2001]
Options:
A) $ -\frac{1}{1+x^{2}} $
B) $ \frac{1}{1+a^{2}}-\frac{1}{1+x^{2}} $
C) $ \frac{1}{1+{{( \frac{a-x}{1+ax} )}^{2}}} $
D) $ \frac{-1}{\sqrt{1-{{( \frac{a-x}{1+ax} )}^{2}}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{d}{dx}[ {{\tan }^{-1}}( \frac{a-x}{1+ax} ) ]. $
= $ \frac{d}{dx}[{{\tan }^{-1}}a-{{\tan }^{-1}}x]=0-\frac{1}{1+x^{2}}=-\frac{1}{1+x^{2}} $ .
BETA
