This is a question from a sample test. I'm just not quite sure how my professor has gotten to this answer. I'm good up until the part where she starts differentiating
$\mathcal{L}${x$^2$y''}
= $\frac{d^2}{ds^2}$[s$^2$$\mathcal{L}${y} - sy(0) - y'(0)]
= $\frac{d}{ds}$[2s$\mathcal{L}$ + s$^2$$\frac{d\mathcal{L}{(y)}}{ds}$ - y(0)]
=2$\mathcal{L}${y} + 4s$\frac{d\mathcal{L}(y)}{ds}$ + s$^2$$\frac{d^2\mathcal{L}{(y)}}{ds^2}$
She applied the product rule: $$(fg)'=f'g+fg'$$ So that: $$\dfrac {d}{ds}(s^2 \mathcal{ L}(y))=2s \mathcal{ L}(y)+s^2\dfrac {d}{ds} (\mathcal{ L}(y))$$ Because $\mathcal{ L}(y)$ is also a function of $s$.