Differentiation Question 239
Question: If $ y={{\cot }^{-1}}( \frac{1+x}{1-x} ) $ , then $ \frac{dy}{dx}= $
[DSSE 1984]
Options:
A) $ \frac{1}{1+x^{2}} $
B) $ -\frac{1}{1+x^{2}} $
C) $ \frac{2}{1+x^{2}} $
D) $ -\frac{2}{1+x^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ y={{\cot }^{-1}}( \frac{1+x}{1-x} ) $
$ \frac{dy}{dx}=-\frac{1}{1+{{( \frac{1+x}{1-x} )}^{2}}}[ \frac{(1-x)+(1+x)}{{{(1-x)}^{2}}} ] $
$ =-\frac{2{{(1-x)}^{2}}}{2(1+x^{2})(1-x^{2})}=-\frac{1}{1+x^{2}} $ .
BETA
