Question

How do you differentiate `g(x) = (2e^(x^2) + 4e^x) ( 2x + 2x^2)` using the product rule?

Answer 1

The product rule states that for a function `y=f(x)g(x)`, `(dy)/(dx)=f'(x)g(x)+f(x)g'(x)`.

Explanation:

We'll also need chain rule, which indicates that `(dy)/(dx)=(dy)/(du)(du)/(dx)`

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

`(dg(x))/(dx)=(4e^x(x+1))(2x(x+1))+(2e^x(e^x+2))(2+4x)`

`(dg(x))/(dx)=(x+1)(4e^x+2x)+(2e^x(e^x+2))(2+4x)`