question_answer

Differentiate \[{{\tan }^{-1}}\left( \frac{x}{1+6{{x}^{2}}} \right)\] with respect to x.

Answer:

Let\[y={{\tan }^{-\,1}}\left( \frac{x}{1+6{{x}^{2}}} \right)\] \[={{\tan }^{-\,1}}\left( \frac{3x-2x}{1+(3x)\,\,(2x)} \right)={{\tan }^{-\,1}}3x-{{\tan }^{-\,1}}2x\] Now, differentiating both sides w.r.t. x, we get \[\frac{dy}{dx}=\frac{1}{1+{{(3x)}^{2}}}\cdot \frac{d}{dx}(3x)-\frac{1}{1+{{(2x)}^{2}}}\cdot \frac{d}{dx}(2x)\]            \[=\frac{3}{1+9{{x}^{2}}}-\frac{2}{1+4{{x}^{2}}}\]