Differentiate (1+x^2) with respect to ^ -1 x - Vidyadip

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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