question_answer

Differentiate \[{{\tan }^{-1}}\left( \frac{1+2x}{1-2x} \right)\] with respect to \[\sqrt{1+4{{x}^{2}}.}\]

Answer:

Let \[u={{\tan }^{-1}}\left( \frac{1+2x}{1-2x} \right)\] and \[v=\sqrt{1+4{{x}^{2}}}.\] Then, \[u=ta{{n}^{-1}}1+{{\tan }^{-1}}2x\] and \[v=\sqrt{1+4{{x}^{2}}}\] \[\Rightarrow \] \[\frac{du}{dx}=\frac{2}{1+4{{x}^{2}}}\] and \[\frac{dv}{dx}=\frac{2}{2\sqrt{1+4{{x}^{2}}}}\,8x\]  \[=\frac{4x}{\sqrt{1+4{{x}^{2}}}}\] \[\therefore \] \[\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\frac{2}{1+4{{x}^{2}}}}{\frac{4x}{\sqrt{1+4{{x}^{2}}}}}\] \[=\frac{1}{2x\sqrt{1+4{{x}^{2}}}}\]