# What is the derivative of? :  d/dx (x^2 + 1)^(x)

Nov 28, 2017

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

#### Explanation:

Let:

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

Taking natural logarithms of both sides:

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

And using the properties of logarithms we have:

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

We now differentiate wrt $x$ and apply the product rule:

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

Then using the chain rule this becomes:

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

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

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