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
Correct Answer: B
Solution :
\[\frac{d}{dx}{{\tan }^{-1}}\frac{4\sqrt{x}}{1-4x}\] \[=\frac{1}{1+{{\left( \frac{4\sqrt{x}}{1-4x} \right)}^{2}}}.\left[ \frac{(1-4x)4\left( \frac{1}{2\sqrt{x}} \right)-4\sqrt{x}(-4)}{{{(1-4x)}^{2}}} \right]\] \[=\frac{2(1+4x)}{\sqrt{x}{{(1+4x)}^{2}}}=\frac{2}{\sqrt{x}(1+4x)}\].
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