AP Calculus AB

## Learning Outcomes

- Define higher-order derivatives and use standard notation such as $$f''(x)$$ and $$\frac{d^2y}{dx^2}$$.
- Calculate second and third derivatives from algebraic functions using derivative rules.
- Interpret the second derivative as the rate of change of the first derivative.
- Connect the sign of $$f''$$ with a preview of concavity: concave up and concave down.
- Identify possible points of inflection and explain why a sign change in concavity is needed.
- Find second derivatives for composite and implicitly defined functions.
- Use graphical, numerical, verbal, and analytic information to reason about concavity.

## AP Calculus AB Syllabus

In AP Calculus AB Unit 3, higher-order derivatives appear after you have learned the chain rule, implicit differentiation, and derivatives of inverse functions. The AP focus is not only on computing derivatives, but also on interpreting them. Concavity is formally used more heavily in applications of derivatives, but Unit 3 introduces the key tool: the second derivative.

- Differentiate a first derivative to obtain a second derivative, when it exists.
- Use notation such as $$f''(x)$$, $$y''$$, $$\frac{d^2y}{dx^2}$$, and $$f^{(n)}(x)$$.
- Apply product, quotient, chain, implicit, and inverse-function derivative techniques when finding higher-order derivatives.
- Interpret $$f''(x)$$ as the rate at which $$f'(x)$$ changes.
- Preview concavity by connecting $$f''>0$$ with concave up behavior and $$f''<0$$ with concave down behavior.
- Recognize that functions may be given algebraically, graphically, numerically, or verbally.
- Avoid relying on calculator derivative values at points where differentiability is not guaranteed.

## Test Your Knowledge

Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision.

1. If $$f(x)=x^4-6x^2+3$$, what are $$f'(x)$$ and $$f''(x)$$?
2. What does the sign of $$f''(x)$$ tell you about the graph of $$f$$?
3. If $$f''(a)=0$$, does $$x=a$$ have to be a point of inflection? Explain.
4. For $$x^2+y^2=1$$, why does finding $$\frac{d^2y}{dx^2}$$ require substituting for $$\frac{dy}{dx}$$?
5. If a table shows values of $$f'(x)$$ increasing as $$x$$ increases, what does that suggest about the concavity of $$f$$?

## Introduction

A first derivative tells you how a function is changing. A second derivative tells you how the first derivative is changing. That simple idea is the start of concavity. If the slopes of tangent lines are increasing, the graph bends upward; if the slopes are decreasing, the graph bends downward.

In Unit 3, your main job is to calculate higher-order derivatives accurately. Concavity is a preview of how those calculations will later help you analyze graphs, justify increasing and decreasing behavior, locate inflection points, and solve application problems such as motion.

> **Key Term: higher-order derivative**
> A derivative obtained by differentiating a function more than once. The second derivative is found by differentiating the first derivative; the third derivative is found by differentiating the second derivative.

### From first derivatives to second derivatives

The derivative of $$y=f(x)$$ may be written in several ways:

f'(x),\quad y',\quad \frac{dy}{dx}
A higher-order derivative is found by differentiating again. The second derivative is written as:

f''(x),\quad y'',\quad \frac{d^2y}{dx^2}
The third derivative is written as:

f'''(x),\quad y''',\quad \frac{d^3y}{dx^3}
For still higher derivatives, AP Calculus often uses:

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

where $$n$$ gives the order of the derivative.

> **Key Term: second derivative**
> The derivative of the first derivative. If $$y=f(x)$$, then $$f''(x)=\frac{d}{dx}\left(f'(x)\right)$$.

For example, let:

$$
f(x)=x^5-4x^3+7x
$$

First differentiate:

f'(x)=5x^4-12x^2+7
Then differentiate again:

f''(x)=20x^3-24x
If needed, continue:

f'''(x)=60x^2-24
This process is straightforward when the function is a polynomial. On the AP exam, however, higher-order derivatives often appear after product, quotient, chain, or implicit differentiation. The second derivative may be more algebraically complex than the first.

> **Test Tip:**
> When asked for $$f''(x)$$, do not stop after finding $$f'(x)$$. Many wrong answers on derivative questions are simply first derivatives given when the question asks for second derivatives.

Consider a composite function:

> **Question:**
> Find $$f''(x)$$ if $$f(x)=\sin(3x^2)$$.

First derivative:

f'(x)=\cos(3x^2)\cdot 6x=6x\cos(3x^2)
Now differentiate using the product rule and chain rule:

f''(x)=6\cos(3x^2)+6x\left[-\sin(3x^2)\cdot 6x\right]
So:

f''(x)=6\cos(3x^2)-36x^2\sin(3x^2)
The main AP skill here is selecting the correct derivative procedures. Higher-order derivative questions often test whether you can keep applying the rules after the first derivative.

### Using the second derivative to read concavity

Concavity describes the way a graph bends. This is not the same as increasing or decreasing. A function can be increasing and concave down, or decreasing and concave up. Increasing and decreasing are controlled by $$f'$$. Concavity is controlled by how $$f'$$ changes, which is measured by $$f''$$.

> **Key Term: concavity**
> The bending behavior of a graph, determined by whether the slopes of tangent lines are increasing or decreasing.

If $$f'(x)$$ is increasing on an interval, then $$f$$ is concave up on that interval. If $$f'(x)$$ is decreasing on an interval, then $$f$$ is concave down on that interval.

Because $$f''$$ is the derivative of $$f'$$:

- If $$f''(x)>0$$ on an interval, then $$f'(x)$$ is increasing there, so $$f$$ is concave up.
- If $$f''(x)<0$$ on an interval, then $$f'(x)$$ is decreasing there, so $$f$$ is concave down.

> **Key Term: concave up**
> A graph is concave up on an interval when its slopes are increasing on that interval; commonly, this occurs when $$f''(x)>0$$.
>
> **Key Term: concave down**
> A graph is concave down on an interval when its slopes are decreasing on that interval; commonly, this occurs when $$f''(x)<0$$.

A useful visual idea is this:

- Concave up: the graph bends like a cup opening upward.
- Concave down: the graph bends like a cap opening downward.

But for AP justifications, use derivative language. Say “$$f''(x)>0$$, so $$f$$ is concave up,” or “$$f'$$ is increasing, so $$f$$ is concave up.”

> **Question:**
> Determine where $$f(x)=x^3-6x^2+9x+1$$ is concave up or concave down.

Find the first derivative:

f'(x)=3x^2-12x+9
Find the second derivative:

f''(x)=6x-12
Set the second derivative equal to zero to find where the sign may change:

$$
6x-12=0
x=2
$$

Test intervals:

- If $$x<2$$, then $$f''(x)<0$$, so $$f$$ is concave down.
- If $$x>2$$, then $$f''(x)>0$$, so $$f$$ is concave up.

Thus, the graph is concave down on $$(-\infty,2)$$ and concave up on $$(2,\infty)$$.

This also previews an inflection point at $$x=2$$ because the concavity changes there.

> **Key Term: point of inflection**
> A point on the graph of a function where concavity changes from up to down or from down to up.
>
> **Exam Warning:**
> The condition $$f''(a)=0$$ does not by itself prove an inflection point. You must have a change in concavity.

For example:

$$
f(x)=x^4
$$

Then:

f'(x)=4x^3
f''(x)=12x^2
At $$x=0$$:

f''(0)=0
But $$12x^2>0$$ for all $$x\ne 0$$, so the graph is concave up on both sides of $$0$$. There is no concavity change, so $$x=0$$ is not an inflection point.

### Finding concavity from formulas, graphs, and tables

AP Calculus AB questions may represent derivatives in different ways. You might be given a formula for $$f$$, a formula for $$f'$$, a graph of $$f'$$, a table of values, or a verbal description.

If you are given $$f(x)$$, find $$f''(x)$$ and analyze its sign.

> **Question:**
> Find the intervals of concavity for $$g(x)=xe^x$$.

First derivative using the product rule:

g'(x)=e^x+xe^x=e^x(1+x)
Second derivative:

g''(x)=e^x(1+x)+e^x=e^x(x+2)
Since $$e^x>0$$ for all real $$x$$, the sign of $$g''(x)$$ depends on $$x+2$$.

- For $$x<-2$$, $$g''(x)<0$$, so $$g$$ is concave down.
- For $$x>-2$$, $$g''(x)>0$$, so $$g$$ is concave up.

There is a concavity change at $$x=-2$$.

If you are given $$f'(x)$$, you do not need to find $$f(x)$$. Concavity comes from whether $$f'$$ is increasing or decreasing. You may also differentiate $$f'$$ to get $$f''$$.

For example, if:

f'(x)=\ln(x^2+1)
then:

f''(x)=\frac{2x}{x^2+1}
Since the denominator is always positive, $$f''(x)<0$$ when $$x<0$$ and $$f''(x)>0$$ when $$x>0$$. Thus, $$f$$ is concave down on $$(-\infty,0)$$ and concave up on $$(0,\infty)$$.

If you are given the graph of $$f'$$, read concavity from the slope of that graph:

- Where the graph of $$f'$$ rises, $$f'$$ is increasing, so $$f$$ is concave up.
- Where the graph of $$f'$$ falls, $$f'$$ is decreasing, so $$f$$ is concave down.

If you are given a table of $$f'$$ values, compare the values as $$x$$ increases. Suppose a table shows:

f'(1)=2,\quad f'(2)=5,\quad f'(3)=9
The values of $$f'$$ are increasing, suggesting $$f$$ is concave up over that interval. If the table shows:

f'(1)=9,\quad f'(2)=5,\quad f'(3)=2
then $$f'$$ is decreasing, suggesting $$f$$ is concave down.

With tables, you usually give a cautious conclusion based on the given data unless the problem states extra information, such as differentiability or monotonic behavior between data points.

> **Test Tip:**
> If the graph shown is $$f'$$, not $$f$$, do not describe concavity by looking for “cup” or “cap” shapes on that graph. Concavity of $$f$$ depends on whether the graph of $$f'$$ is increasing or decreasing.

Concavity also appears in motion. If $$s(t)$$ is position, then:

v(t)=s'(t)
and:

a(t)=v'(t)=s''(t)
So the second derivative of position is acceleration. If $$s''(t)>0$$, velocity is increasing. If $$s''(t)<0$$, velocity is decreasing. This is another way to interpret concavity: position is concave up when velocity increases and concave down when velocity decreases.

### Implicit and composite functions: second derivative examples

Higher-order derivatives can be found for implicitly defined functions as well as explicitly defined functions. In implicit differentiation, treat $$y$$ as a function of $$x$$. Every time you differentiate a term involving $$y$$ with respect to $$x$$, use the chain rule and include $$\frac{dy}{dx}$$.

For second derivatives, you often first find $$\frac{dy}{dx}$$, then differentiate that expression again.

> **Question:**
> For $$x^2+y^2=1$$, find $$\frac{dy}{dx}$$ and $$\frac{d^2y}{dx^2}$$.

Differentiate both sides with respect to $$x$$:

$$
2x+2y\frac{dy}{dx}=0
$$

Solve for $$\frac{dy}{dx}$$:

$$
\frac{dy}{dx}=-\frac{x}{y}
$$

Now differentiate again:

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

Use the quotient rule:

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

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

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

Simplify:

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

Since $$x^2+y^2=1$$:

$$
\frac{d^2y}{dx^2}=-\frac{1}{y^3}
$$

This example shows two common AP features. First, second derivatives in implicit problems may contain both $$x$$ and $$y$$. Second, simplification often uses the original equation.

> **Exam Warning:**
> In implicit second derivative problems, do not leave an unsimplified expression containing $$\frac{dy}{dx}$$ if you can substitute the first derivative. AP questions often expect $$\frac{d^2y}{dx^2}$$ in terms of $$x$$ and $$y$$ only.

Here is another example involving a product and chain rule.

> **Question:**
> Find $$y''$$ if $$y=x^2\cos x$$.

First derivative:

y'=2x\cos x-x^2\sin x
Second derivative:

y''=\frac{d}{dx}(2x\cos x)-\frac{d}{dx}(x^2\sin x)
Use the product rule twice:

$$
\frac{d}{dx}(2x\cos x)=2\cos x-2x\sin x
\frac{d}{dx}(x^2\sin x)=2x\sin x+x^2\cos x
$$

So:

y''=2\cos x-2x\sin x-\left(2x\sin x+x^2\cos x\right)
y''=2\cos x-4x\sin x-x^2\cos x
This derivative can now be used to discuss concavity, although solving the sign of that expression may be difficult without technology or a graph. On AP questions, if exact sign analysis is too difficult, the question may provide a calculator-permitted setting, a graph, or enough data to support a conclusion.

You should also understand that derivatives may fail to exist. Concavity may change at places where $$f''=0$$ or where $$f''$$ does not exist, but only if the function is defined there and the concavity actually changes.

For instance, consider:

$$
f(x)=x^{1/3}
$$

Then:

f'(x)=\frac{1}{3}x^{-2/3}
and:

f''(x)=-\frac{2}{9}x^{-5/3}
The second derivative is not defined at $$x=0$$. For $$x<0$$, $$x^{5/3}<0$$, so $$f''(x)>0$$. For $$x>0$$, $$f''(x)<0$$. Concavity changes from up to down at $$x=0$$, and the function is defined there, so there is an inflection point at $$x=0$$.

Compare that with a vertical asymptote or a point not in the domain. A function cannot have an inflection point at an $$x$$-value where the graph has no point.

Calculator use can help, but it must be used carefully. Numerical derivative commands can be misleading near sharp corners, cusps, vertical tangents, or discontinuities. A calculator may return a number even when a derivative does not exist because it is estimating using nearby values.

> **Test Tip:**
> Before using a calculator derivative value as evidence, check that the function is differentiable at the point. For concavity, also check that the second derivative exists or that the graph/data clearly supports a change in the behavior of $$f'$$.

## Key Point Checklist

This article has covered the following key knowledge points:

- Higher-order derivatives are found by differentiating repeatedly.
- The second derivative of $$f$$ is the derivative of $$f'$$.
- Common second derivative notation includes $$f''(x)$$, $$y''$$, and $$\frac{d^2y}{dx^2}$$.
- The notation $$f^{(n)}(x)$$ means the $$n$$th derivative of $$f$$.
- Concavity describes how a graph bends, not whether it rises or falls.
- If $$f''(x)>0$$ on an interval, then $$f$$ is concave up on that interval.
- If $$f''(x)<0$$ on an interval, then $$f$$ is concave down on that interval.
- If $$f'$$ is increasing, $$f$$ is concave up; if $$f'$$ is decreasing, $$f$$ is concave down.
- A possible point of inflection occurs where $$f''=0$$ or $$f''$$ does not exist, but a concavity change is required.
- For implicit second derivatives, substitute the expression for $$\frac{dy}{dx}$$ after differentiating again.
- A graph of $$f'$$ gives concavity of $$f$$ through whether $$f'$$ is increasing or decreasing.
- Calculator derivative values should not replace checking differentiability and sign changes.

## Key Terms and Concepts

- higher-order derivative
- second derivative
- concavity
- concave up
- concave down
- point of inflection


Calculating higher-order derivatives - Concavity preview