Question

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

Answer 1

`xe^{15x} ( 2+15x)`

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*g` is (Leibniz formula) `f'*g + f*g'`;
• The derivative of a power `x^n` is `nx^{n-1}`;
• The derivative of a composite function `f(g(x))` is `f'(g(x)) * g'(x)`;
• The derivative of an exponential `e^x` is the exponential itself.

Let's put all these things together:

• The derivative of `x^2 e^{15x}` is

`(d/dx x^2) * e^{15x} + x^2 * (d/dx e^{15x})`

• The derivative of `x^2 is`2x^1=2x#, and the expression becomes

`2x * e^{15x} + x^2 * (d/dx e^{15x})`

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

`d/dx e^{15x} = e^{15x} * (d/dx 15x) = e^{15x} * 15`

So, the answer is

`2x * e^{15x} + 15x^2 e^{15x}`

We can factor an exponential and a `x`, obtaining

`xe^{15x} ( 2+15x)`