Differentiation Question 449
Question: If $ y=\log {{( \frac{1+x}{1-x} )}^{1/4}}-\frac{1}{2}{{\tan }^{-1}}x, $ then $ \frac{dy}{dx}= $
Options:
A) $ \frac{x^{2}}{1-x^{4}} $
B) $ \frac{2x^{2}}{1-x^{4}} $
C) $ \frac{x^{2}}{2(1-x^{4})} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=\log {{( \frac{1+x}{1-x} )}^{1/4}}-\frac{1}{2}{{\tan }^{-1}}x $
Differentiating w.r.t. x of y, we get $ \frac{dy}{dx}={{( \frac{1-x}{1+x} )}^{1/4}}\frac{1}{4}{{( \frac{1+x}{1-x} )}^{-3/4}}[ \frac{(1-x)+(1+x)}{{{(1-x)}^{2}}} ]-\frac{1}{2}.\frac{1}{1+x^{2}} $
$ =\frac{1}{2}( \frac{1-x}{1+x} )\frac{1}{{{(1-x)}^{2}}}-\frac{1}{2(1+x^{2})} $
$ =\frac{1}{2}.\frac{1}{(1-x^{2})}-\frac{1}{2}\frac{1}{(1+x^{2})}=\frac{x^{2}}{1-x^{4}} $ .
BETA
