Math Core

The derivative of a function is itself a function, so you can differentiate it again. The result, the second derivative, tells you how the slope is changing. It measures acceleration in motion problems and, in Unit 5, the bending (concavity) of a graph. This lesson is about computing second and higher derivatives quickly and accurately, including for curves defined implicitly.

Notation

Start with $y = f(x)$ . Differentiating repeatedly gives:

derivative	prime notation	Leibniz notation
first	$f'(x)$ or $y'$	$\dfrac{dy}{dx}$
second	$f''(x)$ or $y''$	$\dfrac{d^2y}{dx^2}$
third	$f'''(x)$ or $y'''$	$\dfrac{d^3y}{dx^3}$
$n$ th	$f^{(n)}(x)$	$\dfrac{d^ny}{dx^n}$

After three primes, switch to a number in parentheses: $f^{(4)}(x)$ is the fourth derivative. The parentheses matter, since $f^4(x)$ could be read as a power.

The Leibniz symbol $\dfrac{d^2y}{dx^2}$ comes from applying $\dfrac{d}{dx}$ twice: $\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$ .

Definition

Second derivative

The second derivative of $f$ is the derivative of $f'$ :

f''(x) = \frac{d}{dx}\big(f'(x)\big).

It gives the instantaneous rate of change of the slope of $f$ .

Seeing the three graphs together

For $f(x) = x^3 - 3x$ , the first derivative is $f'(x) = 3x^2 - 3$ and the second is $f''(x) = 6x$ . In the graph below, notice that $f'$ is zero exactly where $f$ has a horizontal tangent ( $x = \pm 1$ ), and $f''$ is zero where the slope of $f$ stops decreasing and starts increasing ( $x = 0$ ).

f(x) = x³ − 3x, its first derivative 3x² − 3, and its second derivative 6x.Open in grapher →

Computing higher derivatives

There's nothing new to learn: apply your rules again, simplifying between steps so the next derivative is easier.

Worked example: Polynomials and powers

Find $f''(x)$ and $f'''(x)$ for $f(x) = 2x^4 - 5x^3 + x - 8$ .

Solution.

\begin{aligned} f'(x) &= 8x^3 - 15x^2 + 1 \\ f''(x) &= 24x^2 - 30x \\ f'''(x) &= 48x - 30. \end{aligned}

Each derivative lowers the degree of a polynomial by one, so $f^{(5)}(x) = 0$ for this degree-4 polynomial.

Worked example: Chain rule, twice

Find $\dfrac{d^2y}{dx^2}$ if $y = \sin(3x)$ .

Solution. $\dfrac{dy}{dx} = 3\cos(3x)$ . Differentiate again, using the chain rule again:

\frac{d^2y}{dx^2} = 3 \cdot \big(-\sin(3x)\big) \cdot 3 = -9\sin(3x).

Notice that $\dfrac{d^2y}{dx^2} = -9y$ . Each derivative brings out another factor of 3.

Worked example: Finding a pattern

Find $f^{(35)}(x)$ for $f(x) = \cos x$ .

Solution. The derivatives of cosine cycle with period 4:

\cos x \;\to\; -\sin x \;\to\; -\cos x \;\to\; \sin x \;\to\; \cos x \;\to\; \cdots

Since $35 = 4 \cdot 8 + 3$ , the 35th derivative matches the 3rd: $f^{(35)}(x) = \sin x$ .

Second derivatives of implicit curves

For an implicitly defined curve, $\dfrac{dy}{dx}$ usually contains $y$ . When you differentiate it again, every $y$ produces another $\dfrac{dy}{dx}$ , which you then replace with the expression you already found. This is a classic AP free-response task.

Implicit second derivatives

Find $\dfrac{dy}{dx}$ by implicit differentiation.
Differentiate $\dfrac{dy}{dx}$ with respect to $x$ (usually with the quotient rule), writing $\dfrac{dy}{dx}$ wherever a $y$ is differentiated.
Substitute your expression for $\dfrac{dy}{dx}$ and simplify. Use the original equation if it helps.

If you only need a value at a point, you can plug in numbers right after step 2.

Worked example: Second derivative on a circle

For $x^2 + y^2 = 25$ , find $\dfrac{d^2y}{dx^2}$ in terms of $y$ , then evaluate it at $(3, 4)$ .

Solution. From the implicit differentiation lesson, $\dfrac{dy}{dx} = -\dfrac{x}{y}$ . Use the quotient rule:

\begin{aligned} \frac{d^2y}{dx^2} &= -\frac{y \cdot 1 - x\dfrac{dy}{dx}}{y^2} = -\frac{y - x\left(-\dfrac{x}{y}\right)}{y^2} \\ &= -\frac{y + \dfrac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3} = -\frac{25}{y^3}. \end{aligned}

The last step used the original equation, $x^2 + y^2 = 25$ . At $(3, 4)$ : $\dfrac{d^2y}{dx^2} = -\dfrac{25}{64}$ .

Common mistake

When differentiating $-\dfrac{x}{y}$ , don't treat $y$ as a constant. The derivative of $y$ with respect to $x$ is $\dfrac{dy}{dx}$ , not 0. Also, $\dfrac{d^2y}{dx^2}$ is not $\left(\dfrac{dy}{dx}\right)^2$ .

Tip

To check an implicit second derivative, pick a point where you can also solve for $y$ explicitly. On the upper half of the circle, $y = \sqrt{25 - x^2}$ ; differentiating twice and plugging in $x = 3$ gives $-\dfrac{25}{64}$ too.

Practice

Practice 1

Find $f''(x)$ for $f(x) = 4x^3 - 5x^2 + 7x$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 2

Let $f(x) = \sqrt{x}$ . Find $f''(4)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find $\dfrac{d^2y}{dx^2}$ if $y = e^{-3x}$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 4

Find $\dfrac{d^3y}{dx^3}$ if $y = \ln x$ .

Enter an expression, e.g. 3x^2 - 2x + 1

Practice 5

(AP-style) If $f(x) = xe^x$ , then $f''(x) =$

Practice 6

(AP-style) If $f(x) = \sin x$ , what is $f^{(27)}(x)$ ?

Practice 7

The point $(5, 4)$ lies on the curve $x^2 - y^2 = 9$ . Find the value of $\dfrac{d^2y}{dx^2}$ at $(5, 4)$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.