Question

If `f(x) = -x^2 -2x` and `g(x) = e^(x)`, what is `f'(g(x))`?

Answer 1

`f'[g(x)] = -2e^x(e^x+1)`

Explanation:

`f(x)=-x^2-2x`
`g(x) = e^x`

Replacing `x` by `g(x)` in `f(x)`

`f[g(x)] = -(e^x)^2 -2*e^x`

`f' [g(x)] = d/dx(-(e^x)^2 -2*e^x) = = d/dx(-e^(2x) -2*e^x)`

`= -2e^(2x)-2e^x`

`= -2e^x(e^x+1)`