# How do I find the derivative of y = log (x^2 + 1)?

Jun 8, 2017

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

#### Explanation:

The derivative of $y = \log \left(u\right)$ is $\frac{\mathrm{dy}}{\mathrm{dx}} = \log e \cdot \frac{1}{u} \times \frac{d}{\mathrm{dx}} \left(u\right)$, since $\ln x = \frac{1}{\log e} \log x$, and $\frac{d \left(\ln x\right)}{\mathrm{dx}} = \frac{1}{x}$.

Following these exact steps we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \log e \cdot \frac{1}{{x}^{2} + 1} \times 2 x$

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

Jun 9, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{\left({x}^{2} + 1\right) \ln 10}$

#### Explanation:

$y = \log \left({x}^{2} + 1\right)$

We can't differentiate $\log \left({x}^{2} + 1\right)$, so we must rewrite in such a way that we can. As such, the change of base rule will be required:

${\log}_{a} b \equiv {\log}_{c} \frac{b}{\log} _ c a$

$\log \left({x}^{2} + 1\right) \equiv \ln \frac{{x}^{2} + 1}{\ln} 10$

$y = \ln \frac{{x}^{2} + 1}{\ln} 10$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} 10$ $\left[\frac{d}{\mathrm{dx}} \left(\ln \left({x}^{2} + 1\right)\right)\right]$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} 10 \times \frac{2 x}{{x}^{2} + 1} = \frac{2 x}{\left({x}^{2} + 1\right) \ln 10}$