A) \[\frac{1}{2(1+x)\,\sqrt{x}}\]
B) \[\frac{3}{(1+x)\,\sqrt{x}}\]
C) \[\frac{2}{(1+x)\,\sqrt{x}}\]
D) \[\frac{2\sqrt{2}y-3}{2\sqrt{2}}=\frac{-3\sqrt{2}\times +3}{2\sqrt{2}}\]
E) \[\frac{3}{2(1+x)\sqrt{x}}\]
Correct Answer: E
Solution :
\[\frac{d}{dx}\,\left( {{\tan }^{-1}}\frac{(\sqrt{x}(3-x)}{1-3x} \right)\] Put \[\sqrt{x}=\tan \theta \Rightarrow \theta ={{\tan }^{-1}}\sqrt{x}\] \[\frac{d}{dx}\left( {{\tan }^{-1}}\frac{(\tan \theta (3-{{\tan }^{2}}\theta )}{1-3{{\tan }^{2}}\theta } \right)\] \[\frac{d}{dx}\left( {{\tan }^{-1}}\frac{(3\tan \theta -{{\tan }^{3}}\theta )}{1-3{{\tan }^{2}}\theta } \right)\] \[\frac{d}{dx}({{\tan }^{-1}}(\tan 3\theta )=\frac{d}{dx}(3\theta )\] \[\frac{d}{dx}(3.{{\tan }^{-1}}\sqrt{x})=\frac{3}{2\sqrt{x}(1+x)}\].
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