# How do you find the derivative of e^(1+lnx)?

Jun 4, 2017

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

#### Explanation:

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

First, rewrite the expression using ${a}^{b + c} \equiv {a}^{b} \cdot {a}^{c}$

${e}^{1 + \ln x} = {e}^{1} \cdot {e}^{\ln} x = x e$

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

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