Differentiation Question 236
Question: If $ y={{\tan }^{-1}}( \frac{{x^{1/3}}+{a^{1/3}}}{1-{x^{1/3}}{a^{1/3}}} ) $ , then $ \frac{dy}{dx}= $
[DSSE 1986]
Options:
A) $ \frac{1}{3{x^{2/3}}(1+{x^{2/3}})} $
B) $ \frac{1}{3{x^{2/3}}(1+{x^{2/3}})} $
C) $ -\frac{1}{3{x^{2/3}}(1+{x^{2/3}})} $
D) $ -\frac{a}{3{x^{2/3}}(1+{x^{2/3}})} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{\tan }^{-1}}( \frac{{x^{1/3}}+{a^{1/3}}}{1-{x^{1/3}}.{a^{1/3}}} )={{\tan }^{-1}}({x^{1/3}})+{{\tan }^{-1}}{a^{1/3}} $
Therefore $ \frac{dy}{dx}=\frac{1}{3{x^{2/3}}(1+{x^{2/3}})} $ .
BETA
