# How do you find the derivative of ln(x^2+1)?

Dec 11, 2016

$\frac{d}{\mathrm{dx}} \ln \left({x}^{2} + 1\right) = \frac{2 x}{{x}^{2} + 1}$

#### Explanation:

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If $y = f \left(x\right)$ then $f ' \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

I was taught to remember that the differential can be treated like a fraction and that the "$\mathrm{dx}$'s" of a common variable will "cancel" (It is important to realise that $\frac{\mathrm{dy}}{\mathrm{dx}}$ isn't a fraction but an operator that acts on a function, there is no such thing as "$\mathrm{dx}$" or "$\mathrm{dy}$" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$ etc, or $\left(\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\textcolor{red}{\cancel{\mathrm{dv}}}} \frac{\textcolor{red}{\cancel{\mathrm{dv}}}}{\textcolor{b l u e}{\cancel{\mathrm{du}}}} \frac{\textcolor{b l u e}{\cancel{\mathrm{du}}}}{\mathrm{dx}}\right)$

So with $y = \ln \left({x}^{2} + 1\right)$, Then:

$\left\{\begin{matrix}\text{Let "u=x^2+1 & => & (du)/dx=2x \\ "Then } y = \ln u & \implies & \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}\end{matrix}\right.$

Using $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dy}}{\mathrm{du}}\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$ we get:

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