| Last time: | Next time: |
Consider it a video game, and challenge yourself to become a black belt in this one. It will pay off.
If a function's derivative is a function, does that function have a derivative?
The derivative of the derivative of a function \(f\) is called the second derivative of \(f\), and frequently denoted \(f''\). And how do we interpret these "higher derivatives" in the context of a motion?
\[ s(t)=\frac{1}{2}gt^2+v_0t+h_0 \]
Each of the coefficients has an important, intuitive role to play:
Of course we could continue on and define higher derivatives (third, fourth, etc.). The notation for the $n^{th}$ derivative is frequently $f^{(n)}(x)$, where the parenthese around the "exponent" mean we're to understand that as differentiation: \[ s'''(t)\equiv s^{(3)}(t)=0 \]
(The third derivative is called the "jerk", amusingly enough! There's no jerk in the flight of the eraser. You might feel some jerk if you're on a roller coaster....)
We can continue, of course, and so we see that
$s^{(n)}(t)=0$ for all $n>2$.
\[ f^\prime(x) = \frac{dy}{dx} = \frac{df}{dx} = \frac{d}{dx}\left(f(x)\right)=\left(f(x)\right)'=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \] \[ f^{\prime\prime}(x) = (f^\prime(x))^\prime = f^{(2)}(x)=\frac{d^2f}{dx^2}=\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{d}{dx}\left( f(x) \right) \right) =\lim_{h \to 0} \frac{f'(x+h)-f'(x)}{h} \]
Several of these forms remind us that differentiation is itself a function (we call it an "operator", since it works on functions): it takes a function in its domain and returns another function in its range -- the derivative function.
But most importantly for us, we realize that the second derivative can be thought of as the derivative of the derivative of \(f\): \[ f''(x)=\lim_{h\to 0}{\frac{f'(x+h)-f'(x)}{h}} \] Hence, all your table work (forward, backward, centered difference formulas) still applies, it's just that we use values for \(f'\) to do our calculations.
This is our poster child for inflection: