Question

Let h(x) = e^(kx) + f(x),

Answer 1

$\bf h(x)=e^{kx}+f(x)\implies \cfrac{dh}{dx}=ke^{kx}+f'(x) \\\\\\ \begin{cases} x=0\\ f'(0)=-8 \end{cases} h'(0)=ke^{k(0)}+(-8)\implies h'(0)=k-8$