# How do you find the derivative of y= e^(ex) ln x?

Feb 5, 2017

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

#### Explanation:

Using the product rule:

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

Now we have using the chain rule:

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

while:

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

is a standard derivative.

Substituting this in (1) we have:

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

and simplifying:

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