Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation3 Marks
Question
Differentiate $\log(1+\text{x}^2)$ with respect to $\tan^{-1}\text{x}$
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Answer
Let $\text{u}=\log(1+\text{x}^2)$ Differentiating it with respect to x using chain rule, $\frac{\text{du}}{\text{dx}}=\frac{1}{(1+\text{x}^2)}\frac{\text{d}}{\text{dx}}(1+\text{x}^2)$ $=\frac{1}{(1+\text{x}^2)}(2\text{x})$ $\frac{\text{du}}{\text{dx}}=\frac{2\text{x}}{(1+\text{x}^2)}\ .....(\text{i})$ Let $\text{v}=\tan^{-1}\text{x}$ Differentiating it with respect to x, $\frac{\text{dv}}{\text{dx}}=\frac{1}{1+\text{x}^2}\ .....(\text{ii})$ Dividing equation (i) by (ii), $\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2\text{x}}{(1+\text{x}^2)}\times\frac{(1+\text{x}^2)}{1}$ $\frac{\text{du}}{\text{dx}}=2\text{x}$
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