Differentiation Question 150
Question: $ \frac{d}{dx}{{\tan }^{-1}}\frac{4\sqrt{x}}{1-4x}= $
Options:
A) $ \frac{1}{\sqrt{x}(1+4x)} $
B) $ \frac{2}{\sqrt{x}(1+4x)} $
C) $ \frac{4}{\sqrt{x}(1+4x)} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{d}{dx}{{\tan }^{-1}}\frac{4\sqrt{x}}{1-4x} $
$ =\frac{1}{1+{{( \frac{4\sqrt{x}}{1-4x} )}^{2}}}.[ \frac{(1-4x)4( \frac{1}{2\sqrt{x}} )-4\sqrt{x}(-4)}{{{(1-4x)}^{2}}} ] $
$ =\frac{2(1+4x)}{\sqrt{x}{{(1+4x)}^{2}}}=\frac{2}{\sqrt{x}(1+4x)} $ .
BETA
