# What are the critical values, if any, of f(x) = f(x) = x^{2}e^{15 x}?

Nov 4, 2015

$x {e}^{15 x} \left(2 + 15 x\right)$

#### Explanation:

To find the critical points, we need the first derivative. This function is a multiplication of a power and a composite exponential. Let's see how to deal with these three things:

• The derivative of a multiplication $f \cdot g$ is (Leibniz formula) $f ' \cdot g + f \cdot g '$;
• The derivative of a power ${x}^{n}$ is $n {x}^{n - 1}$;
• The derivative of a composite function $f \left(g \left(x\right)\right)$ is $f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$;
• The derivative of an exponential ${e}^{x}$ is the exponential itself.

Let's put all these things together:

• The derivative of ${x}^{2} {e}^{15 x}$ is

$\left(\frac{d}{\mathrm{dx}} {x}^{2}\right) \cdot {e}^{15 x} + {x}^{2} \cdot \left(\frac{d}{\mathrm{dx}} {e}^{15 x}\right)$

• The derivative of ${x}^{2} i s$2x^1=2x#, and the expression becomes

$2 x \cdot {e}^{15 x} + {x}^{2} \cdot \left(\frac{d}{\mathrm{dx}} {e}^{15 x}\right)$

• The exponential is a composite function, so we must derive the exponential and then multiply for the derivative of the exponent:

$\frac{d}{\mathrm{dx}} {e}^{15 x} = {e}^{15 x} \cdot \left(\frac{d}{\mathrm{dx}} 15 x\right) = {e}^{15 x} \cdot 15$

$2 x \cdot {e}^{15 x} + 15 {x}^{2} {e}^{15 x}$
We can factor an exponential and a $x$, obtaining
$x {e}^{15 x} \left(2 + 15 x\right)$