---
title: "AP Calculus 3.6: Calculating Higher-Order Derivatives"
description: "Review AP Calculus Topic 3.6, including higher-order derivatives, second derivatives, third derivatives, nth derivative notation, repeated differentiation, concavity, acceleration, and worked examples."
canonical: "https://fiveable.me/ap-calc/unit-3/calculating-higher-order-derivatives/study-guide/Mh7ZnLES3ycIzEEKWl95"
type: "study-guide"
subject: "AP Calculus AB/BC"
unit: "Unit 3 – Composite, Implicit, and Inverse Functions"
lastUpdated: "2026-06-09"
---

# AP Calculus 3.6: Calculating Higher-Order Derivatives

## Summary

Review AP Calculus Topic 3.6, including higher-order derivatives, second derivatives, third derivatives, nth derivative notation, repeated differentiation, concavity, acceleration, and worked examples.

## Guide

A higher-order [derivative](/ap-calc/unit-10-infinite-sequences-and-series-bc-only-/finding-taylor-polynomial-approximations-functions/study-guide/LszguYzKz0M6GdqTRSr6 "fv-autolink") just means you keep differentiating: take the derivative of the [first derivative](/ap-calc/key-terms/first-derivative "fv-autolink") to get the second, then differentiate again for the third, and so on. The process is the same as finding a first derivative; you just apply your rules again to the result you already have. For AP Calculus, second derivatives often connect to concavity, acceleration, and graph behavior.

## Why This Matters for the AP Calculus Exam

Higher-order derivatives show up across the AP Calculus exam because the second derivative carries a lot of meaning. The second derivative tells you about concavity and [points of inflection](/ap-calc/key-terms/points-of-inflection "fv-autolink"), and in motion problems it connects position, velocity, and acceleration. On multiple-choice and free-response questions, you may need to compute a second derivative, read it from notation, or use it to interpret a function's behavior. Getting comfortable with repeated differentiation here sets you up for [Unit 4](/ap-calc/unit-4 "fv-autolink") motion problems and Unit 5 analysis of functions.

## Key Takeaways

- A higher-order derivative comes from differentiating a derivative: differentiate $f'$ to get $f''$, differentiate $f''$ to get $f'''$, and so on, as long as each derivative exists.
- Each step uses the same differentiation rules you already know (power, product, [quotient](/ap-calc/unit-1/determining-limits-using-algebraic-properties-limits/study-guide/HjStgVKViPGZj1CxYwEB "fv-autolink"), chain, trig).
- Know the notation: the second derivative can be $f''(x)$, $y''$, or $\frac{d^2y}{dx^2}$; a general $n$th derivative is $f^{(n)}(x)$ or $\frac{d^ny}{dx^n}$.
- The second derivative gives concavity and points of inflection; in motion it gives acceleration.
- For polynomials, repeated differentiation eventually reaches a constant, then 0.
- Simplify carefully between steps so the next derivative is easier to take.

## Higher-Order Derivatives Explained

A higher-order derivative just means you take the derivative more than once. After the first derivative $f'(x)$, you can keep going:

- The **second derivative** is the derivative of $f'(x)$.
- The **third derivative** is the derivative of $f''(x)$.
- The **$n$th derivative** repeats this process $n$ times.

Each derivative only exists if the one before it is [differentiable](/ap-calc/key-terms/differentiable "fv-autolink"), so the chain of derivatives can stop if a function is not smooth enough.

### Why the Second Derivative Is Useful

The first and second derivatives tell you a lot about a graph:

- The **first derivative** $f'(x)$ helps locate relative minimums and maximums, since those often occur where the [slope](/ap-calc/key-terms/slope "fv-autolink") is zero.
- The **second derivative** $f''(x)$ describes **concavity** and helps you find **points of inflection**, where $f(x)$ switches from [concave up](/ap-calc/key-terms/concave-up "fv-autolink") to [concave down](/ap-calc/key-terms/concave-down "fv-autolink") or the other way.

### Notation You Need to Know

The second derivative can be written as:

$$
f''(x),\quad y'',\quad \frac{d^2y}{dx^2}
$$

Higher-order derivatives can be written as:

$$
f^{(n)}(x),\quad \frac{d^ny}{dx^n}
$$

The parentheses in $f^{(n)}(x)$ matter: it means the $n$th derivative, not $f$ raised to a power.

## Calculating Higher-Order Derivatives Step by Step

The big idea: finding a second, third, or $n$th derivative follows the same process as the first derivative. You just apply your rules to the previous result.

Take this function:

$$
f(x)=\frac{2}{3}x^3+4x^2+3x-1
$$

Use the [Power Rule](/ap-calc/unit-2/applying-power-rule/study-guide/GMr6EEbZezsP1DvqrpEk) for the first derivative:

$$
f'(x)=2x^2+8x+3
$$

The second derivative is the derivative of $f'(x)$:

$$
f''(x)=4x+8
$$

Keep going for the third derivative (the derivative of $f''(x)$):

$$
f'''(x)=4
$$

$$
f^{(4)}(x)=0
$$

Once you reach a constant, the next derivative is 0. For polynomials, you will typically be asked for the first, second, or third derivative.

## Practice Problems

For each function $f(x)$, find the second derivative $f''(x)$.

### Set 1: Quick Power Rules and Trig

**Example 1:**

$$
f(x)=6x^4-2x^2+5x+1
$$

First derivative using the [Power Rule](/ap-calc/unit-2/applying-power-rule/study-guide/GMr6EEbZezsP1DvqrpEk "fv-autolink"):

$$
f'(x)=24x^3-4x+5
$$

Apply the Power Rule again for the second derivative:

$$
f''(x)=72x^2-4
$$

**Example 2:**

$$
f(x)=\sin(x)
$$

Trig derivatives are ones you should memorize. If you need a refresher, see the guide on [Trig Function Derivatives](/ap-calc/unit-2/derivatives-cos-x-sinx-e-x-ln-x/study-guide/SbmDK3t2kYI0u2wLz9Hv).

The derivative of $\sin(x)$ is $\cos(x)$:

$$
f'(x)=\cos(x)
$$

The derivative of $\cos(x)$ is $-\sin(x)$, so:

$$
f''(x)=-\sin(x)
$$

**Example 3:**

$$
f(x)=\cos(2x)
$$

This is a **[composite function](/ap-calc/key-terms/composite-function "fv-autolink")**: a function inside another function. The outer function is $\cos(x)$ and the inner function is $2x$, so you need the [Chain Rule](/ap-calc/unit-3/chain-rule/study-guide/27HxeRGCYJBjuPWBm1uw):

$$
\frac{d}{dx}[O(I(x))]=O'(I(x))\cdot I'(x)
$$

Set things up with outer function $O$ and inner function $I$:

$$
O(x)=\cos(x) \rightarrow O'(x)=-\sin(x)
$$

$$
I(x)=2x\rightarrow I'(x)=2
$$

First derivative:

$$
f'(x)=-\sin(2x)\cdot 2=-2\sin(2x)
$$

Repeat for the second derivative, now with $-2\sin(x)$ as the outer function:

$$
O'(x)=-2\cos(x),\quad I'(x)=2
$$

$$
f''(x)=-2\cos(2x)\cdot 2=-4\cos(2x)
$$

### Set 2: Chain and Product Rule

**Example 4:**

$$
f(x)=(5x^4+2x^2-3x+9)^2
$$

The outer function is $x^2$ and the inner function is the polynomial inside the parentheses. Use [the Chain Rule](/ap-calc/unit-3/chain-rule/study-guide/27HxeRGCYJBjuPWBm1uw "fv-autolink"):

$$
f'(x)=O'(I(x))\cdot I'(x)
$$

$$
O(x) = x^2 \rightarrow O'(x)=2x
$$

$$
I(x)=5x^4+2x^2-3x+9\rightarrow I'(x)=20x^3+4x-3
$$

$$
f'(x)=2(5x^4+2x^2-3x+9)\cdot(20x^3+4x-3)
$$

$$
f'(x)=(10x^4+4x^2-6x+18)\cdot(20x^3+4x-3)
$$

Instead of multiplying everything out, use the [Product Rule](/ap-calc/unit-2/product-rule/study-guide/qQXYTmpHvjsqAWOVzcEw) on the two factors, calling them $L(x)$ and $R(x)$:

$$
(L(x)\cdot R(x))' = L'(x)\cdot R(x) + L(x)\cdot R'(x)
$$

$$
L(x) = 10x^4+4x^2-6x+18\rightarrow L'(x)=40x^3+8x-6
$$

$$
R(x)=20x^3+4x-3\rightarrow R'(x)=60x^2+4
$$

Putting it together:

$$
f''(x)=(40x^3+8x-6)(20x^3+4x-3)+(10x^4+4x^2-6x+18)(60x^2+4)
$$

You can leave it like this, or expand for practice:

$$
f''(x)=1400x^6+600x^4-600x^3+1128x^2-72x+90
$$

**Example 5:**

$$
f(x)=\sqrt{5x^3+81x^2}
$$

Rewrite the square root as a power so it is easier to differentiate:

$$
f(x)=(5x^3+81x^2)^\frac{1}{2}
$$

Now apply the Chain Rule:

$$
O(x)=x^\frac{1}{2}\rightarrow O'(x)=\frac{1}{2}x^{-\frac{1}{2}}
$$

$$
I(x)=5x^3+81x^2\rightarrow I'(x)=15x^2+162x
$$

$$
f'(x)=\frac{1}{2}\left(5x^3+81x^2\right)^{-\frac{1}{2}}\cdot \left(15x^2+162x\right)
$$

For the second derivative, use [the Product Rule](/ap-calc/unit-2/product-rule/study-guide/qQXYTmpHvjsqAWOVzcEw "fv-autolink"). The left factor needs the Chain Rule again; the right factor uses the Power Rule:

$$
L(x)=\frac{1}{2}(5x^3+81x^2)^{-\frac{1}{2}}\rightarrow L'(x)=\frac{-1}{4}(5x^3+81x^2)^{-\frac{3}{2}}\cdot(15x^2+162)
$$

$$
R(x)=15x^2+162x \rightarrow R'(x)=30x+162
$$

Combining gives:

$$
f''(x)=L'(x)\cdot R(x) + L(x)\cdot R'(x)
$$

$$
f''(x)=\frac{15(5x+108)}{4(5x+81)^\frac{3}{2}}
$$

Leaving an answer unsimplified is fine; the main goal is choosing the right rules for the type of function.

### Set 3: Rational Functions and Natural Logs

**Example 6:**

$$
f(x)=\tan(3x)+\ln(x)
$$

The derivative of $\tan(x)$ is $\sec^2(x)$, and the derivative of $\ln(x)$ is $\frac{1}{x}$. Apply the Chain Rule to the first term:

$$
f'(x)=\sec^2(3x)\cdot 3+\frac{1}{x}=3\sec^2(3x)+\frac{1}{x}
$$

Rewrite $\frac{1}{x}$ as $x^{-1}$ to make the next step cleaner. The derivative of $\sec(x)$ is $\sec(x)\tan(x)$, so applying the Chain and Power Rules gives:

$$
f''(x)=18\sec^3(3x)\tan(3x)-\frac{1}{x^2}
$$

**Example 7:**

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

This is a quotient, so use the [Quotient Rule](/ap-calc/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc):

$$
\frac{d}{dx}\left[\frac{N(x)}{D(x)}\right]=\frac{D(x)N'(x)-N(x)D'(x)}{(D(x))^2}
$$

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

Apply [the Quotient Rule](/ap-calc/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc "fv-autolink") again (with the Chain Rule on the denominator) for the second derivative:

$$
f''(x)=\frac{(2x+2)(x+1)^2-2(x+1)(x^2+2x)}{((x+1)^2)^2}=\frac{2}{(x+1)^3}
$$

## How to Use This on the AP Calculus Exam

### MCQ

- Be ready to compute a second derivative quickly, especially for polynomials and basic trig functions.
- Watch the notation. $f^{(4)}(x)$ means the fourth derivative, not $f(x)$ to the fourth power.
- If a question asks about concavity or inflection points, that is a signal to use the second derivative.

### Free Response

- Show each derivative step clearly so it is easy to follow your work to $f''(x)$.
- Simplify the first derivative before differentiating again. A cleaner $f'(x)$ makes the second derivative much easier.
- When a function has a product, quotient, or composite, identify which rule applies before you start, then reapply that rule for the next derivative.

### Common Trap

- Forgetting to apply the Chain Rule again on higher-order derivatives of composite functions like $\cos(2x)$ or $\sec^2(3x)$.

## Common Misconceptions

- **"Higher-order derivatives need new rules."** They do not. You use the same power, product, quotient, chain, and trig rules, just applied to the previous derivative.
- **"$f^{(n)}(x)$ means $f(x)$ raised to the $n$th power."** The parentheses around $n$ mean the $n$th derivative, not an exponent.
- **"The second derivative tells you maximums and minimums directly."** The first derivative locates where slope is zero; the second derivative tells you about concavity and inflection points.
- **"Once you take the derivative, you can drop the Chain Rule."** Composite functions still need the Chain Rule every time you differentiate, including for the second and third derivatives.
- **"Polynomial derivatives never end."** For a polynomial, repeated differentiation reaches a constant and then becomes 0.

## Related AP Calculus Guides

- [3.2 Implicit Differentiation](/ap-calc/unit-3/implicit-differentiation/study-guide/k43S7kJDyGg9NFUm78Uw)
- [3.1 The Chain Rule](/ap-calc/unit-3/chain-rule/study-guide/27HxeRGCYJBjuPWBm1uw)
- [3.3 Differentiating Inverse Functions](/ap-calc/unit-3/differentiating-inverse-functions/study-guide/u6wuCCW76Syq9wzKdkKD)
- [3.4 Differentiating Inverse Trigonometric Functions](/ap-calc/unit-3/differentiating-inverse-trigonometric-functions/study-guide/pzDRhb3oXKntPRoQgLOz)
- [3.5 Selecting Procedures for Calculating Derivatives](/ap-calc/unit-3/selecting-procedures-for-calculating-derivatives/study-guide/8zbglfs22PQH1ZvWssHd)
- [Unit 3 Overview: Differentiation: Composite, Implicit, and Inverse Functions](/ap-calc/unit-3/review/study-guide/2ugbv0Bkh0SQ2yTAgf2j)

## Vocabulary

- **first derivative**: The derivative of a function, denoted f', which describes the rate of change and indicates where a function is increasing or decreasing.
- **higher-order derivatives**: Derivatives of derivatives obtained by repeatedly differentiating a function; the second derivative, third derivative, and beyond.
- **second derivative**: The derivative of the first derivative, denoted f'', which describes the concavity of a function and indicates where it is concave up or concave down.

## FAQs

### What are higher-order derivatives?

Higher-order derivatives come from differentiating repeatedly: the derivative of f' is f'', the derivative of f'' is f''', and the process continues as long as each derivative exists.

### What does the second derivative tell you?

The second derivative describes how the first derivative changes. In graph analysis, it helps determine concavity and points of inflection; in motion, it represents acceleration.

### How do you write higher-order derivative notation?

Common notation includes f''(x), y'', d^2y/dx^2 for the second derivative, and f^(n)(x) or d^ny/dx^n for the nth derivative.

### How do you calculate a second derivative?

First find f'(x), then differentiate that result to get f''(x). Simplify between steps when it makes the next derivative easier.

### What happens to higher derivatives of a polynomial?

For a polynomial, repeated differentiation eventually reaches a constant, and the next derivative is zero.

### Where do higher-order derivatives appear on the AP Calculus exam?

They appear in motion, concavity, inflection point, graph analysis, and procedural derivative questions, especially when a second derivative is needed.

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