Laplace transform of $tf'(t)$

Question

I know that $\mathcal{L}(tf'(t)) = -\frac{d}{ds}\mathcal{L}(f'(t))$ and that this $= -\frac{d}{ds}(sF(s) - f(0))$ but the solution says that this becomes $-F(s) - F'(s)$ and I can't figure out why $\frac{d}{ds}f(0) = sF'(s)$. Shouldn't it be $0$ ?

Answer 1

Notice that $$-\frac{d}{ds}(sF(s)-f(0))=-\frac{d}{ds}sF(s)+\frac{d}{ds}f(0)=-F(s)-sF'(s)+0=-F(s)-sF'(s)$$