Differentiation Question 127
Question: Differential coefficient of $ \frac{{{\tan }^{-1}}x}{1+{{\tan }^{-1}}x} $ w.r.t. $ {{\tan }^{-1}}x $ is
Options:
A) $ \frac{1}{1+{{\tan }^{-1}}x} $
B) $ \frac{-1}{1+{{\tan }^{-1}}x} $
C) $ \frac{1}{{{(1+{{\tan }^{-1}}x)}^{^{2}}}} $
D) $ \frac{-1}{2{{(1+{{\tan }^{-1}}x)}^{2}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
The differential coefficient of $ \frac{{{\tan }^{-1}}x}{1+{{\tan }^{-1}}x} $ with respect to $ {{\tan }^{-1}}x=\frac{\frac{d}{dx}( \frac{{{\tan }^{-1}}x}{1+{{\tan }^{-1}}x} )}{\frac{d}{dx}({{\tan }^{-1}}x)} $
$ =\frac{(1+{{\tan }^{-1}}x)\frac{d}{dx}{{\tan }^{-1}}x-{{\tan }^{-1}}x\frac{d}{dx}(1+{{\tan }^{-1}}x)}{{{(1+{{\tan }^{-1}}x)}^{2}}.\frac{d}{dx}{{\tan }^{-1}}x} $
$ =\frac{1}{{{(1+{{\tan }^{-1}}x)}^{2}}} $ .
BETA
