Question

`f(x) =-3e^(x+4)+e^-3` Find `f'(x)` ?

Answer 1

`f'(x)= -3e^4`
[Assuming `f(x) =-3e^(x+4)+e^-3`]

Explanation:

Since we have no exponentiation or parentheses in the function as written (in the pre-restoration question), I will take it as:

`f(x) =-3e^(x+4)+e^-3`

Apply chain rule

`f'(x) = -3e^(x+4) * d/dx (x+4) +0`

`= -3e^(x+4) *1 = -3e^(x+4)`

`f'(0)= -3e^4`

NB: If we take the function as written in the question:

`f(x) = -3ex +4 +e-3`

Where all but the first term are constants.

Hence, `f'(x) = -3e forall x in RR`

`:.f'(0) = -3e`