Derivatives — AP Calculus AB Calc AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: The derivative as instantaneous rate of change, core differentiation rules (power, product, quotient, chain), implicit differentiation, derivatives of exponential, logarithmic and trigonometric functions, and higher-order derivatives.

You should already know: Strong precalculus (functions, trig, algebra).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Derivatives?

The derivative of a function $f(x)$ at a point $x=a$ is a measure of how steeply the function’s output changes as its input changes at that exact location, equivalent to the slope of the tangent line to the graph of $f$ at $x=a$. Common notations for derivatives include $f^{\prime }(a)$, $\frac{dy}{dx}{|}_{x=a}$, $\frac{d}{dx}\left[f(x)\right]{|}_{x=a}$, and $y^{\prime }$. Synonyms for the derivative include instantaneous rate of change, tangent slope, and marginal value (in economics applications). Per the AP Calculus AB CED, derivatives make up 17-20% of the total exam score, appearing in both multiple-choice and free-response sections.

2. Derivative as instantaneous rate of change

To understand the derivative as an instantaneous rate of change, first recall the average rate of change of $f$ over the interval $[a,a+h]$ is given by the slope of the secant line connecting the points $(a,f(a))$ and $(a+h,f(a+h))$: $$\text{Average rate of change}=\frac{f(a+h)-f(a)}{h}$$ As $h$ approaches 0, the two points on the secant line move closer together, and the secant converges to the tangent line at $x=a$. The limit of the average rate of change as $h\to 0$ is the instantaneous rate of change, i.e., the derivative: $$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$$ An alternative equivalent form of the limit definition is: $$f^{\prime }(a)=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}$$

Worked Example

Use the limit definition to find $f^{\prime }(2)$ for $f(x)=3x^2+2x$.

1. Compute $f(2+h)=3(2+h{)}^2+2(2+h)=3(4+4h+h^2)+4+2h=16+14h+3h^2$
2. Compute $f(2)=3(2{)}^2+2(2)=16$
3. Substitute into the limit definition: $\frac{f(2+h)-f(2)}{h}=\frac{14h+3h^2}{h}=14+3h$ for $h\ne 0$
4. Evaluate the limit as $h\to 0$: $f^{\prime }(2)=14+0=14$ This means at $x=2$, $f(x)$ is increasing at a rate of 14 units per unit increase in $x$.

Exam tip: Examiners regularly test recognition of the limit definition of a derivative. If you see a limit matching the forms above, it is always the derivative of $f$ at the given point $a$.

3. Differentiation rules — power, product, quotient, chain

The limit definition is accurate but tedious for complex functions. The following standard rules let you compute derivatives directly without evaluating limits, and appear on every AP Calculus AB exam:

1. Power Rule: For $f(x)=x^n$ where $n$ is any real constant: $$f^{\prime }(x)=n{x}^{n-1}$$ Works for integer, fractional, and negative exponents: $\frac{d}{dx}[x^5]=5x^4$, $\frac{d}{dx}[\sqrt{x}]=\frac{1}{2}{x}^{-1/2}$, $\frac{d}{dx}\left[\frac{1}{x^3}\right]=-3x^{-4}$.
2. Product Rule: For $f(x)=u(x)\cdot v(x)$: $$f^{\prime }(x)=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$ Mnemonic: "first derivative of second plus second derivative of first". Example: For $f(x)=(2x+1)(x^2-3)$, $u=2x+1,u^{\prime }=2$, $v=x^2-3,v^{\prime }=2x$, so $f^{\prime }(x)=2(x^2-3)+(2x+1)(2x)=6x^2+2x-6$.
3. Quotient Rule: For $f(x)=\frac{u(x)}{v(x)}$, $v(x)\ne 0$: $$f^{\prime }(x)=\frac{u^{\prime }(x)v(x)-u(x)v^{\prime }(x)}{{\left[v(x)\right]}^2}$$ Mnemonic: "low d high minus high d low, over the square of what's below". Example: For $f(x)=\frac{3x+2}{x-1}$, $f^{\prime }(x)=\frac{3(x-1)-(3x+2)(1)}{(x-1{)}^2}=\frac{-5}{(x-1{)}^2}$.
4. Chain Rule: For composite functions $f(x)=g(h(x))$: $$f^{\prime }(x)=g^{\prime }(h(x))\cdot h^{\prime }(x)$$ In plain terms: differentiate the outer function (keeping the inner function unchanged), then multiply by the derivative of the inner function. Example: For $f(x)=(2x^3+5{)}^4$, outer $g(u)=u^4,g^{\prime }(u)=4u^3$, inner $h(x)=2x^3+5,h^{\prime }(x)=6x^2$, so $f^{\prime }(x)=4(2x^3+5{)}^3\cdot 6x^2=24x^2(2x^3+5{)}^3$.

Exam tip: The chain rule is the most frequently tested differentiation rule on AP Calc AB, often combined with other rules. Always check for composite functions before finalizing your derivative.

4. Implicit differentiation

All functions differentiated so far are explicit, meaning $y$ is isolated on one side of the equation ($y=f(x)$). Implicit functions have $x$ and $y$ mixed together (e.g., $x^2+y^2=25$, the equation of a circle, which cannot be written as a single explicit function of $x$). Implicit differentiation lets you find $\frac{dy}{dx}$ without solving for $y$ first.

The process follows three steps:

1. Differentiate both sides of the equation with respect to $x$
2. Every time you differentiate a term containing $y$, multiply by $\frac{dy}{dx}$ (by the chain rule, since $y$ is a function of $x$)
3. Rearrange to isolate $\frac{dy}{dx}$

Worked Example

Find $\frac{dy}{dx}$ for the implicit function $3xy+y^3=2x$.

1. Differentiate both sides with respect to $x$: $$\frac{d}{dx}[3xy]+\frac{d}{dx}[y^3]=\frac{d}{dx}[2x]$$
2. Apply the product rule to $3xy$ and chain rule to $y^3$: $$3\left(y+x\frac{dy}{dx}\right)+3y^2\frac{dy}{dx}=2$$
3. Expand and collect terms with $\frac{dy}{dx}$: $$3y+3x\frac{dy}{dx}+3y^2\frac{dy}{dx}=2⟹\frac{dy}{dx}(3x+3y^2)=2-3y$$
4. Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{2-3y}{3x+3y^2}$$

Exam tip: Examiners frequently ask for the slope of a tangent line to an implicit curve at a given point. Plug in the $(x,y)$ coordinates immediately after solving for $\frac{dy}{dx}$ to avoid unnecessary algebraic simplification.

5. Derivatives of $e^x,\ln x$, trig functions

These core transcendental function derivatives are guaranteed to appear on the AP Calculus AB exam, and all assume angles are measured in radians:

1. Exponential functions: The derivative of $e^x$ is itself, a unique property of the natural exponential function. Apply the chain rule for composite exponents: $$\frac{d}{dx}[e^x]=e^x\frac{d}{dx}\left[e^{u(x)}\right]=e^{u(x)}\cdot u^{\prime }(x)$$ Example: $\frac{d}{dx}\left[e^{3x^2}\right]=e^{3x^2}\cdot 6x=6x{e}^{3x^2}$
2. Logarithmic functions: The derivative of the natural log $\ln x$ is the reciprocal of $x$, valid for $x>0$: $$\frac{d}{dx}[\ln x]=\frac{1}{x}\frac{d}{dx}\left[\ln (u(x))\right]=\frac{u^{\prime }(x)}{u(x)},u(x)>0$$ Example: $\frac{d}{dx}[\ln (2x+5)]=\frac{2}{2x+5}$
3. Trigonometric functions: Derivatives of co-functions (cos, csc, cot) have a negative sign: $$\frac{d}{dx}[\sin x]=\cos x\frac{d}{dx}[\cos x]=-\sin x\frac{d}{dx}[\tan x]={\sec }^{2}x$$ $$\frac{d}{dx}[\csc x]=-\csc x\cot x\frac{d}{dx}[\sec x]=\sec x\tan x\frac{d}{dx}[\cot x]=-{\csc }^{2}x$$ Example: $\frac{d}{dx}[\sin (4x)]=\cos (4x)\cdot 4=4\cos (4x)$

Exam tip: A common trap is forgetting the chain rule for trigonometric functions with a linear argument (e.g., $\sin (2x)$). Always multiply by the derivative of the angle term.

6. Higher-order derivatives

The first derivative measures the rate of change of the original function, but you can differentiate derivatives themselves to get higher-order derivatives, which measure the rate of change of the rate of change. Standard notation for higher-order derivatives is:

• 1st derivative: $f^{\prime }(x)$, $\frac{dy}{dx}$, $y^{\prime }$
• 2nd derivative: Differentiate the 1st derivative: $f^{\prime \prime }(x)$, $\frac{d^{2}y}{d{x}^2}$, $y^{\prime \prime }$
• 3rd derivative: Differentiate the 2nd derivative: $f^{\prime \prime \prime }(x)$, $\frac{d^{3}y}{d{x}^3}$, $y^{\prime \prime \prime }$
• For orders $n\ge 4$, use parenthetical notation: $f^{(n)}(x)$, $\frac{d^{n}y}{d{x}^n}$, $y^{(n)}$

For AP Calculus AB, you only need to compute up to the 2nd derivative for most applications (concavity, acceleration), but you may be asked for higher orders for simple functions.

Worked Example

For $f(x)=5x^3+2e^{2x}+\sin x$, compute $f^{\prime \prime }(x)$ and $f^{\prime \prime \prime }(0)$.

1. 1st derivative: $f^{\prime }(x)=15x^2+4e^{2x}+\cos x$
2. 2nd derivative: $f^{\prime \prime }(x)=30x+8e^{2x}-\sin x$
3. 3rd derivative: $f^{\prime \prime \prime }(x)=30+16e^{2x}-\cos x$
4. Evaluate at $x=0$: $f^{\prime \prime \prime }(0)=30+16(1)-1=45$

7. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting the chain rule for composite functions, e.g., writing $\frac{d}{dx}[(2x+1{)}^3]=3(2x+1{)}^2$ instead of $3(2x+1{)}^2\cdot 2$. Why: Students rush and only differentiate the outer function, ignoring the inner term. Correct move: Explicitly identify inner and outer functions for every composite term before differentiating, then add the chain rule multiplication step immediately.
• Wrong move: Mixing up the order of terms in the quotient rule, leading to sign errors. Why: Students misremember the mnemonic or swap numerator terms. Correct move: Say the "low d high minus high d low" mnemonic out loud every time you apply the quotient rule, and spot-check your result with a test point if possible.
• Wrong move: Using degree mode when computing numerical values of trigonometric derivatives. Why: Students forget all calculus trig derivative rules are derived using radian measure. Correct move: Set your calculator to radian mode at the start of the exam and never switch it; the AP exam will never ask for degree-based derivative calculations.
• Wrong move: Forgetting to multiply by $\frac{dy}{dx}$ when differentiating $y$ terms in implicit differentiation. Why: Students treat $y$ as a constant instead of a function of $x$. Correct move: Add a $\frac{dy}{dx}$ factor immediately after differentiating any term containing $y$, before moving to the next term.
• Wrong move: Applying the power rule to exponential functions, e.g., writing $\frac{d}{dx}[2^x]=x{2}^{x-1}$. Why: Students confuse power functions (variable base, constant exponent) with exponential functions (constant base, variable exponent). Correct move: Use the power rule only for variable-base terms; for exponential functions, use $\frac{d}{dx}[a^x]=a^x\ln a$ (the natural exponential $e^x$ is the most commonly tested on AB).

8. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

The limit $$\underset{h\to 0}{\lim }\frac{\ln (e^2+h)-2}{h}$$ is the derivative of which function at which point? A) $f(x)=\ln x$ at $x=e^2$ B) $f(x)=\ln x$ at $x=2$ C) $f(x)=e^x$ at $x=2$ D) $f(x)=\ln (e^x)$ at $x=e^2$

Worked Solution: Compare the given limit to the definition $f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$. We see $f(a+h)=\ln (e^2+h)$, so $f(x)=\ln x$ and $a=e^2$. The constant term $2=\ln (e^2)=f(a)$, so the limit is $f^{\prime }(e^2)$ for $f(x)=\ln x$. The correct answer is A.

Question 2 (Free Response Part A)

Consider the curve defined implicitly by $x^2+2xy+y^3=4$. (a) Find $\frac{dy}{dx}$. (b) Find the slope of the tangent line to the curve at the point $(1,1)$.

Worked Solution: (a) Differentiate both sides with respect to $x$: $$2x+2\left(y+x\frac{dy}{dx}\right)+3y^2\frac{dy}{dx}=0$$ Collect terms with $\frac{dy}{dx}$: $$\frac{dy}{dx}(2x+3y^2)=-2x-2y$$ Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{-2x-2y}{2x+3y^2}$$ (b) Substitute $x=1,y=1$: $$\frac{dy}{dx}{|}_{(1,1)}=\frac{-2(1)-2(1)}{2(1)+3(1{)}^2}=\frac{-4}{5}$$ The slope of the tangent line is $-\frac{4}{5}$.

Question 3 (Free Response)

The position of a particle moving along a straight line is given by $s(t)=2t^3-9t^2+12t+5$, where $s$ is measured in meters and $t$ in seconds. (a) Find the velocity function $v(t)$ (velocity is the first derivative of position). (b) Find the acceleration function $a(t)$ (acceleration is the second derivative of position). (c) What is the acceleration of the particle at $t=2$ seconds?

Worked Solution: (a) $v(t)=s^{\prime }(t)=6t^2-18t+12$ m/s (b) $a(t)=v^{\prime }(t)=12t-18$ m/s² (c) Evaluate $a(2)=12(2)-18=6$ m/s²

9. Quick Reference Cheatsheet

10. What's Next

Derivatives are the foundation for 60% of the remaining AP Calculus AB syllabus, so mastering these rules is non-negotiable for success. Next, you will apply derivatives to analyze function behavior: finding increasing/decreasing intervals, local maxima and minima, concavity and inflection points, as well as solving real-world problems including related rates, optimization, and motion along a line. Derivatives also form the basis for the Fundamental Theorem of Calculus, which connects derivatives and integrals, the second core topic of the course.

If you are stuck on any differentiation rule, need extra practice with implicit differentiation, or want to test your knowledge with more exam-style questions, you can ask Ollie, our AI tutor, for personalized help at any time. You can also find more study guides, practice tests, and exam tips on the homepage to prepare for your AP Calculus AB exam.