Answer
a.
$\frac{d^2}{dx^2} [cf(x)] = \frac{d}{dx} (\frac{d}{dx}cf(x)) = \frac{d}{dx} c(\frac{d}{dx}f(x))$
Let $g(x) = \frac{d}{dx}f(x)$.
Then, $\frac{d^2}{dx^2} [cf(x)] = \frac{d}{dx} c(\frac{d}{dx}f(x)) = \frac{d}{dx}c*g(x) = c\frac{d}{dx}g(x) = c\frac{d^2}{dx^2}f(x)$
$\frac{d^2}{dx^2} [f(x) + g(x)] = \frac{d}{dx} (\frac{d}{dx}[f(x)+g(x)]) = \frac{d}{dx} (\frac{d}{dx}f(x) + \frac{d}{dx}g(x))$
Let $h(x) = \frac{d}{dx}f(x)$ and $i(x) = \frac{d}{dx}g(x)$.
Thus, $\frac{d^2}{dx^2} [f(x) + g(x)] = \frac{d}{dx} (\frac{d}{dx}f(x) + \frac{d}{dx}g(x)) = \frac{d}{dx} (h(x) + i(x)) = \frac{d}{dx}h(x) + \frac{d}{dx}i(x) = \frac{d^2}{dx^2}f(x) + \frac{d^2}{dx^2}g(x)$
Work Step by Step
b. The results generalize to the $n$th derivatives as we may continue to the derivates of functions for new functions and continue to move the coefficient out of the current derivative or distribute the derivative. Thus, the results generalize to the $n$th derivative.
