# How do you differentiate y=ln(1+x^2)?

Sep 4, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{1 + {x}^{2}}$

#### Explanation:

differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

That is $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

let $u = 1 + {x}^{2} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 2 x$

and so $y = \ln u \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$

substitute these values into (A) changing u back to terms of x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \left(2 x\right) = \frac{2 x}{1 + {x}^{2}}$