Second and Higher Derivatives — AP® Calculus AB Notes

The derivative $f'$ of a function is a new function. Nothing stops us from differentiating it again. The result, the second derivative $f''$ , tells how fast the rate of change is itself changing. It is the key to motion problems in the next unit and to the shape of graphs after that, so it is worth getting the mechanics and the meaning right now.

Notation

All of these mean the same second derivative:

$f''(x) = y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

Read $\frac{d^2y}{dx^2}$ as "d two y, d x squared". The 2s sit where they do because the operation $\frac{d}{dx}$ is applied twice to $y$ . It is not $\left(\frac{dy}{dx}\right)^2$ , the square of the slope, and not the derivative of $y^2$ . Higher derivatives follow the same pattern: $f'''(x)$ for the third, then $f^{(4)}(x)$ , $f^{(5)}(x)$ and in general $f^{(n)}(x)$ , with the number in brackets so it is not mistaken for a power.

Computing higher derivatives

Differentiate, then differentiate the result. Every rule still applies at every step.

Worked example 1. Find $f''(x)$ and $f'''(x)$ for $f(x) = x^3 - 3x$ .

$f'(x) = 3x^2 - 3$ , then $f''(x) = 6x$ , then $f'''(x) = 6$ , and $f^{(4)}(x) = 0$ .

A polynomial of degree $n$ has $f^{(n+1)} = 0$ : every derivative lowers the degree by one.

Three stacked graphs on the same x-axis: f(x) = x³ − 3x (crimson) with level points at x = −1 and 1; f′(x) = 3x² − 3 (blue), zero at x = ±1; and f″(x) = 6x (green), the slope of f′, negative left of 0 and positive right of 0. — **Figure 1.** $f(x) = x^3 - 3x$ , $f'(x) = 3x^2 - 3$ and $f''(x) = 6x$ . Each graph records the slopes of the one above it. LevlPrep original diagram.

Look at the figure from bottom to top. Where $f''$ is negative ( $x < 0$ ), the graph of $f'$ is falling. Where $f''$ is positive, $f'$ is rising. And where $f'$ is 0, at $x = \pm 1$ , the graph of $f$ levels off. Each function describes the slope of the one above it.

Worked example 2. Find $\frac{d^2y}{dx^2}$ for $y = e^{x^2}$ .

First derivative, by the chain rule: $\frac{dy}{dx} = 2xe^{x^2}$ .

This is a product of $2x$ and $e^{x^2}$ , so the second derivative needs the product rule (and the chain rule again inside it):

$\frac{d^2y}{dx^2}$ $= 2 \cdot e^{x^2}$ $+ 2x \cdot 2xe^{x^2}$ $= (2 + 4x^2)e^{x^2}$ .

The most common slip is to differentiate only one factor of $2xe^{x^2}$ .

Patterns are worth spotting. For $f(x) = \sin(3x)$ : $f' = 3\cos(3x)$ , $f'' = -9\sin(3x)$ , $f''' = -27\cos(3x)$ , and every two derivatives multiply by $-9$ . For $f(x) = e^{-2x}$ , each derivative multiplies by $-2$ , so $f^{(n)}(x)$ $= (-2)^n e^{-2x}$ .

Reading $f''$ from a graph of $f'$

When the exam shows the graph of $f'$ , the second derivative is the slope of that graph, not its height. If the graph of $f'$ is a line segment from $(0, 2)$ to $(2, -2)$ , then $f''(1)$ $= \frac{-2 - 2}{2 - 0}$ $= -2$ , while $f'(1) = 0$ is the height. At a sharp corner of the graph of $f'$ , the slopes from the two sides differ, so $f''$ does not exist there.

On the graph of $f'$	What it tells
height at $x = a$	$f'(a)$ , the slope of $f$
slope at $x = a$	$f''(a)$
graph rising	$f'' > 0$
graph falling	$f'' < 0$
sharp corner	$f''$ does not exist there

Second derivatives of implicit curves

For a curve given implicitly, differentiate the expression for $\frac{dy}{dx}$ again, remembering that $y$ still depends on $x$ . Then substitute the expression for $\frac{dy}{dx}$ wherever it appears.

Worked example 3. For $x^2 + 4y^2 = 8$ , find $\frac{d^2y}{dx^2}$ at $(2, 1)$ .

First: $2x + 8y\frac{dy}{dx} = 0$ , so $\frac{dy}{dx} = -\frac{x}{4y}$ . At $(2, 1)$ , $\frac{dy}{dx} = -\frac{1}{2}$ .

Second, by the quotient rule: $\frac{d^2y}{dx^2}$ $= -\frac{(1)(4y) - x\left(4\frac{dy}{dx}\right)}{(4y)^2}$ .

At $(2, 1)$ with $\frac{dy}{dx} = -\frac{1}{2}$ : $-\frac{4 - 2 \cdot 4 \cdot \left(-\frac{1}{2}\right)}{16}$ $= -\frac{4 + 4}{16}$ $= -\frac{1}{2}$ .

In general, substituting $\frac{dy}{dx} = -\frac{x}{4y}$ and using $x^2 + 4y^2 = 8$ gives $\frac{d^2y}{dx^2}$ $= -\frac{x^2 + 4y^2}{16y^3}$ $= -\frac{1}{2y^3}$ .

Slips that cost points

Writing $\frac{dy^2}{dx}$ or $\left(\frac{dy}{dx}\right)^2$ for the second derivative.
Forgetting that a first derivative such as $2xe^{x^2}$ is a product.
Treating $y$ as a constant when differentiating $\frac{dy}{dx}$ for an implicit curve.
Reading the height of a graph of $f'$ when the question asks for $f''$ .

Notation

Computing higher derivatives

Reading f″f'' from a graph of f′f'

Second derivatives of implicit curves

Slips that cost points

Reading $f''$ from a graph of $f'$