Question

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

Answer 1

Apply the product rule to the functions `f_1(x) = 2x^2 - 4x + 4` and `f_2(x) = e^x` to obtain `d/dxf_1(x)f_2(x) = 2x^2e^x`

Explanation:

The product rule states that for any functions `f(x)` and `g(x)`
`d/dxf(x)g(x) = f'(x)g(x) + f(x)g'(x)`

Let `f_1(x) = 2x^2 - 4x + 4` and `f_2(x) = e^x`

Then `f_1'(x) = 4x - 4` and `f_2'(x) = e^x`

Note that

`d/dx(2x^2 - 4x + 4)e^x = d/dxf_1(x)f_2(x)`

So, applying the product rule gives us

`d/dx(2x^2 - 4x + 4)e^x = f_1'(x)f_2(x) + f_1(x)f_2'(x)`
`=>d/dx(2x^2 - 4x + 4)e^x=(4x-4)e^x+(2x^2-4x+4)e^x`

Simplifying the result gives us the final answer of
`d/dx(2x^2 - 4x + 4)e^x = 2x^2e^x`