Question

How do you find the 1st and 2nd derivative of `x^2e^x`?

Answer 1

`dy/dx = x(x + 2)e^x`
`(d^2y)/(dx^2) = (x^2 +4x + 2)e^x`

Explanation:

By the product rule.

`dy/dx = 2x xx e^x + e^x xx x^2`

`dy/dx = 2xe^x + x^2e^x`

`dy/dx = x(x + 2)e^x`

`(d^2y)/(dx^2) = 2 xx e^x + 2x xx e^x + 2x xx e^x + x^2xxe^x`

`(d^2y)/(dx^2) = 2e^x + 2xe^x + 2xe^x + x^2e^x`

`(d^2y)/(dx^2) = 2e^x + 4xe^x + x^2e^x`

`(d^2y)/(dx^2) = (x^2 + 4x +2)e^x`

Hopefully this helps!