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 \[v=\sqrt{1+4{{x}^{2}}}.\]             Now, \[u={{\tan }^{-1}}1+{{\tan }^{-1}}2x\] and \[v=\sqrt{1+4{{x}^{2}}}\]             \[\Rightarrow \] \[\frac{du}{dx}=\frac{2}{1+4{{x}^{2}}}\] and \[\frac{dv}{dx}=\frac{1}{2\sqrt{1+4{{x}^{2}}}}\cdot \]             \[8x=\frac{4x}{\sqrt{1+4{{x}^{2}}}}\] \[\therefore \] \[\frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{\frac{2}{1+4{{x}^{2}}}}{\frac{4x}{\sqrt{1+4{{x}^{2}}}}}\] \[=\frac{1}{2x\sqrt{1+4{{x}^{2}}}}\]