# What is the derivative of f(x)=e^(x^2lnx)?

Jan 24, 2017

$f ' \left(x\right) = x {e}^{{x}^{2} \ln \left(x\right)} \left(2 \ln \left(x\right) + 1\right)$

#### Explanation:

We will use the following:

• The chain rule.

• The product rule.

• $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

• $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

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

With those:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} {e}^{{x}^{2} \ln \left(x\right)}$

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

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

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

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

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