Laplace transform of piecewise function and its derivative

Question

I have been given this piecewise function F(t) where $$F(t)= \left\{ \begin{array}{ll} 2t & 0\leq t \leq 1 \\ t & t\gt 1 \\ \end{array} \right. $$I have to find its Laplace transform and Laplace transform of its derivative and then show that it satisfies $$L[F'(t)]=sf(s)-F(0) \to (A)$$ where $f(s)=L[F(t)]$. I've tried this as follows: $$f(s)= \int_0^1(2t)e^{-st}dt \, +\int_1^{\infty} t e^{-st}dt\\=\frac {2}{s^2} -\frac{e^{-s}}{s^2} -\frac{e^{-s}}{s}$$ And $$L[F'(t)]=\frac{2}{s}- \frac{e^{-s}}{s}$$ But it doesn't satisfy the relation (A). I've also tried writing function F(t) in terms of Heaviside's unit step function as $$F(t) = 2t-tu(t-1)$$and using this Laplace transform is $$f(s)=\frac {2}{s^2} -\frac{e^{-s}}{s^2} $$ which satisfies the relation (A). My question is that why not the other way? And why it satisfies relation (A) this way? Is there any mistake or some concept involved? I don't know. Thank you in advance.