How do you differentiate f(x)=xe^x-x using the product rule?

Nov 20, 2015

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

Explanation:

Let's first examine just the $x {e}^{x}$ portion of this function.

Through the product rule,
$\frac{d}{\mathrm{dx}} \left[x {e}^{x}\right] = \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[x\right]} \cdot {e}^{x} + \textcolor{p u r p \le}{\frac{d}{\mathrm{dx}} \left[{e}^{x}\right] \cdot} x$

Therefore,
$\frac{d}{\mathrm{dx}} \left[x {e}^{x}\right] = \textcolor{b l u e}{1} \cdot {e}^{x} + \textcolor{p u r p \le}{{e}^{x}} \cdot x = {e}^{x} + x {e}^{x}$

We can use this in the original equation to determine that
f'(x)=e^x+xe^x-d/dx[x]=color(green)(e^x+xe^x-1