Higher-Order Derivatives

The derivative $f'(x)$ is a function too, so you can differentiate it again. The result, $f''(x)$, is the second derivative: it tells you how fast the slope itself is changing. Second derivatives describe acceleration, and they tell you which way a graph bends, which you’ll use a lot in the next units. Any trig on this page is in radians.

Key ideas

Notation

Each new derivative is the derivative of the one before it.

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 $y'''$	$\dfrac{d^3y}{dx^3}$
Fourth	$f^{(4)}(x)$ or $y^{(4)}$	$\dfrac{d^4y}{dx^4}$
$n$th	$f^{(n)}(x)$ or $y^{(n)}$	$\dfrac{d^ny}{dx^n}$

After three primes, switch to a number in brackets: $f^{(4)}$, not $f''''$. The brackets matter, because $f^4$ would look like a power.

In Leibniz notation, $\dfrac{d^2y}{dx^2}$ means $\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$. The $2$ goes on the $d$ on top and on the $x$ on the bottom.

What the second derivative means

• $f(x)$ gives the value.
• $f'(x)$ gives the slope, the rate of change of $f$.
• $f''(x)$ gives the rate of change of the slope.

If $f''(x) \gt 0$, the slope is increasing, and the graph bends upward (concave up). If $f''(x) \lt 0$, the slope is decreasing, and the graph bends downward (concave down). You’ll study this in concavity and the second derivative test.

For motion along a line, if $s(t)$ is position, then $s'(t)$ is velocity and $s''(t)$ is acceleration. See straight-line motion.

Each graph shows the slope of the one above it: $f'$ is zero where $f$ levels off, and $f''$ is zero where $f'$ levels off.

Useful patterns

• A polynomial of degree $n$ becomes a constant after $n$ derivatives, and $0$ after that.
• Sine and cosine cycle every four derivatives: $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$.
• Each derivative of $e^{kx}$ brings out another factor of $k$, so $\dfrac{d^n}{dx^n}e^{kx} = k^n e^{kx}$.

Second derivatives of implicit relations

For a curve defined implicitly:

1. Find $\dfrac{dy}{dx}$ as usual.
2. Differentiate $\dfrac{dy}{dx}$ again with respect to $x$. Remember that $y$ is still a function of $x$, so every $y$ produces a $\dfrac{dy}{dx}$.
3. Substitute your expression for $\dfrac{dy}{dx}$ from step 1.
4. Simplify, often using the original equation.

This kind of question shows up often on the no-calculator part of the AP exam.

Worked examples

Example 1: A polynomial

Find all the nonzero derivatives of $f(x) = 2x^4 - 5x^3 + x - 7$.

Solution. Differentiate one step at a time:

$$\begin{aligned} f'(x) &= 8x^3 - 15x^2 + 1 \\ f''(x) &= 24x^2 - 30x \\ f'''(x) &= 48x - 30 \\ f^{(4)}(x) &= 48 \end{aligned}$$

Every derivative after $f^{(4)}$ is $0$.

Example 2: A trig function

Let $y = \sin(3x)$. Find $\dfrac{d^2y}{dx^2}$, and show that $y'' + 9y = 0$.

Solution. Use the chain rule each time:

$$\frac{dy}{dx} = 3\cos(3x), \qquad \frac{d^2y}{dx^2} = 3 \cdot \big(-\sin(3x)\big) \cdot 3 = -9\sin(3x)$$

Then $y'' + 9y = -9\sin(3x) + 9\sin(3x) = 0$.

Example 3: Product and chain rule

Let $f(x) = xe^{2x}$. Find $f''(x)$ and $f''(0)$.

Solution. Product rule, with the chain rule on $e^{2x}$:

$$f'(x) = 1 \cdot e^{2x} + x \cdot 2e^{2x} = e^{2x}(1 + 2x)$$

Differentiate $f'(x) = e^{2x}(1 + 2x)$ with the product rule again:

$$f''(x) = 2e^{2x}(1 + 2x) + e^{2x} \cdot 2 = e^{2x}(4 + 4x) = 4e^{2x}(1 + x)$$

So $f''(0) = 4e^0(1) = 4$.

Simplifying $f'(x)$ before differentiating again made the second step easier.

Example 4: Second derivative of a circle

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

Solution. From implicit differentiation, $\dfrac{dy}{dx} = -\dfrac{x}{y}$.

Differentiate again with the quotient rule. The derivative of the top, $x$, is $1$; the derivative of the bottom, $y$, is $\dfrac{dy}{dx}$:

$$\frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \dfrac{dy}{dx}}{y^2}$$

Substitute $\dfrac{dy}{dx} = -\dfrac{x}{y}$:

$$\frac{d^2y}{dx^2} = -\frac{y + \dfrac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}$$

From the original equation, $x^2 + y^2 = 25$, so

$$\frac{d^2y}{dx^2} = -\frac{25}{y^3}$$

At $(3, 4)$: $\dfrac{d^2y}{dx^2} = -\dfrac{25}{64}$. It’s negative, which makes sense: the top half of a circle bends downward.

Common mistakes

Squaring the first derivative. $f''(x)$ means “differentiate twice”, not $\big(f'(x)\big)^2$. Likewise $\dfrac{d^2y}{dx^2}$ is not $\left(\dfrac{dy}{dx}\right)^2$.

Plugging in too early. To find $f''(2)$, find the formula for $f''(x)$ first, then substitute. If you substitute $x = 2$ into $f'(x)$ first, you get a number, and the derivative of a number is $0$.

Losing chain-rule factors. $\dfrac{d^2}{dx^2}\sin(3x) = -9\sin(3x)$, not $-3\sin(3x)$ or $-\sin(3x)$. Each differentiation brings out another factor of $3$.

Treating y as a constant in implicit second derivatives. When you differentiate $-\dfrac{x}{y}$, the derivative of $y$ is $\dfrac{dy}{dx}$, not $0$.

Not substituting dy/dx. An implicit second derivative should be written in terms of $x$ and $y$ only. Replace every $\dfrac{dy}{dx}$ with its expression, then simplify.

Writing f⁽⁴⁾ without brackets. $f^4(x)$ looks like $f(x)$ to the fourth power. Use $f^{(4)}(x)$ for the fourth derivative.

Practice

1. (Warm-up) Let $f(x) = x^5 - 4x^2$. Find $f''(x)$ and $f''(1)$.

Solution

$f'(x) = 5x^4 - 8x$, so $f''(x) = 20x^3 - 8$, and $f''(1) = 20 - 8 = 12$.

2. (Warm-up) Let $y = \cos x$. Find $y''$ and $y^{(4)}$.

Solution

$y' = -\sin x$, $y'' = -\cos x$, $y''' = \sin x$, $y^{(4)} = \cos x$.

So $y'' = -\cos x$ and $y^{(4)} = \cos x$ (back where we started).

3. (Warm-up) Let $f(x) = \sqrt{x}$. Find $f''(4)$.

Solution

Write $f(x) = x^{1/2}$:

$$f'(x) = \frac{1}{2}x^{-1/2}, \qquad f''(x) = -\frac{1}{4}x^{-3/2}$$$$f''(4) = -\frac{1}{4} \cdot \frac{1}{4^{3/2}} = -\frac{1}{4} \cdot \frac{1}{8} = -\frac{1}{32}$$

4. (Core) Let $y = e^{-3x}$. Find $\dfrac{d^3y}{dx^3}$, and give a formula for $\dfrac{d^ny}{dx^n}$.

Solution

Each derivative multiplies by $-3$:

$$\frac{dy}{dx} = -3e^{-3x}, \quad \frac{d^2y}{dx^2} = 9e^{-3x}, \quad \frac{d^3y}{dx^3} = -27e^{-3x}$$

In general, $\dfrac{d^ny}{dx^n} = (-3)^n e^{-3x}$.

5. (Core) Let $f(x) = \ln(x^2 + 1)$. Find $f''(x)$, $f''(0)$, and the values of $x$ where $f''(x) = 0$.

Solution

By the chain rule, $f'(x) = \dfrac{2x}{x^2 + 1}$. By the quotient rule:

$$f''(x) = \frac{2(x^2 + 1) - 2x(2x)}{(x^2 + 1)^2} = \frac{2 - 2x^2}{(x^2 + 1)^2}$$

$f''(0) = \dfrac{2}{1} = 2$.

$f''(x) = 0$ when $2 - 2x^2 = 0$, so $x = 1$ or $x = -1$.

6. (Core) A particle moves along a line so that its position is $s(t) = t^3 - 6t^2 + 9t$ metres after $t$ seconds. Find its velocity and acceleration at $t = 1$.

Solution

$$v(t) = s'(t) = 3t^2 - 12t + 9, \qquad a(t) = s''(t) = 6t - 12$$

$v(1) = 3 - 12 + 9 = 0$ m/s and $a(1) = 6 - 12 = -6$ m/s².

At $t = 1$ the particle is momentarily stopped, and its velocity is decreasing.

7. (Core) Let $f(x) = x\sin x$. Find $f''(x)$ and $f''(\pi)$.

Solution

$$f'(x) = \sin x + x\cos x$$$$f''(x) = \cos x + \big(\cos x - x\sin x\big) = 2\cos x - x\sin x$$

$f''(\pi) = 2\cos\pi - \pi\sin\pi = 2(-1) - 0 = -2$.

8. (Challenge) For the hyperbola $x^2 - y^2 = 9$, find $\dfrac{d^2y}{dx^2}$ in terms of $y$, and evaluate it at $(5, 4)$.

Solution

First derivative:

$$2x - 2y\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{x}{y}$$

Second derivative, by the quotient rule:

$$\frac{d^2y}{dx^2} = \frac{y \cdot 1 - x\dfrac{dy}{dx}}{y^2} = \frac{y - \dfrac{x^2}{y}}{y^2} = \frac{y^2 - x^2}{y^3}$$

From the original equation, $y^2 - x^2 = -9$, so

$$\frac{d^2y}{dx^2} = -\frac{9}{y^3}$$

Check $(5, 4)$: $25 - 16 = 9$. Then $\dfrac{d^2y}{dx^2} = -\dfrac{9}{64}$.

9. (Challenge) Let $f(x) = \sin(2x)$. Find $f^{(50)}(x)$.

Solution

Look for the pattern:

$$f'(x) = 2\cos(2x), \quad f''(x) = -4\sin(2x), \quad f'''(x) = -8\cos(2x), \quad f^{(4)}(x) = 16\sin(2x)$$

Each derivative brings out a factor of $2$, and the trig part cycles every $4$ derivatives. Since $50 = 4 \cdot 12 + 2$, the $50$th derivative has the same trig part as $f''$, which is $-\sin(2x)$:

$$f^{(50)}(x) = -2^{50}\sin(2x)$$