Answer
a.) True
b.) False
Work Step by Step
NOTE: $\dfrac{d}{dx}$ indicates deriving once. $\dfrac{d^2}{dx^2}$ indicates deriving twice. So when you see $\dfrac{d}{dx}$, this means we are taking the derivative of a function and are finding the first derivative. Likewise, $\dfrac{d^2}{dx^2}$ means we are finding the second derivative.
a.) We need to prove that $\dfrac{d}{dx}(fg' - f'g) = fg'' - f''g$ In other words we need to prove that the derivative of (fg' - f'g) is equal to fg'' - f''g
so $\dfrac{d}{dx}(fg' - f'g) = (fg'' + f'g') - (f'g' + f''g)$
$(fg' - f'g)' = fg'' + (f'g' - f'g') + f''g = fg'' + 0 + f''g$
$(fg' - f'g)' = fg'' - f''g$
Statement a.) is true
b.) We need to prove that $\dfrac{d^2}{dx^2}(fg) = fg'' + f''g$
so $\dfrac{d}{dx}(fg) = fg' + f'g$
This is the first derivative, now we need to find the second.
so $\dfrac{d^2}{dx^2}(fg) = \dfrac(d)(dx)(fg' + f'g)$
$(fg)'' = (fg'' + f'g') + (f'g' + f''g)$
$(fg)'' = fg'' + 2(f'g') - f''g \ne fg'' + f''g$
statement b.) is false
