Selecting procedures for calculating derivatives — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Chain rule for composite functions, implicit differentiation for implicit relations, derivatives of inverse functions, derivatives of inverse trigonometric functions, and matching appropriate derivative techniques to the form of a given function.

You should already know: Basic derivative rules for power, exponential, trigonometric, and logarithmic functions. The limit definition of the derivative. Product and quotient rules for differentiation.

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 BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Selecting procedures for calculating derivatives?

This is a core skill in Unit 3 (Differentiation: Composite, Implicit, and Inverse Functions) of the AP Calculus BC CED, worth approximately 9–13% of the total AP exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. The skill is not just memorizing derivative rules—it is recognizing the form of the function or relation you are working with, then choosing the correct technique to differentiate efficiently and accurately. Many AP exam problems intentionally mix function forms (e.g., an inverse trigonometric composite function, or an implicit relation with a transcendental term) specifically to test your ability to select the right procedure. You will never be told which rule to use on the exam, so this selection skill is as important as memorizing the rules themselves. Even if you recall all formulas correctly, applying the wrong procedure will always lead to an incorrect result and lost points.

2. Selecting the Chain Rule for Composite Functions

A composite function is written $f(g(x))$, where one function (the inner function $g(x)$) is nested inside another (the outer function $f(u)$ where $u=g(x)$). The chain rule is always the correct procedure for differentiating composite functions. The formula for the chain rule is: $$\frac{d}{dx}\left[f(g(x))\right]=f^{\prime }(g(x))\cdot g^{\prime }(x)$$ In Leibniz notation, this is written $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$, which makes the intuition clear: the total rate of change of $y$ with respect to $x$ is the product of the rate of change of $y$ with respect to the inner function $u$, and the rate of change of $u$ with respect to $x$. You can spot a composite function by checking if you can rewrite it by substituting a new variable for a nested term: common examples include $(2x-\ln x{)}^7$, $\sin (3e^x)$, $e^{\cos x}$, and $\ln (x^2-4x)$. For nested composites (e.g., $e^{\sin (x^2)}$), you apply the chain rule repeatedly, one layer at a time.

Worked Example

Find $\frac{dy}{dx}$ for $y=(2x^3-\cos x{)}^5$.

1. Identify the composite structure: outer function $f(u)=u^5$, inner function $u=2x^3-\cos x$.
2. Differentiate the outer function, leaving the inner function unchanged: $f^{\prime }(u)=5u^4=5(2x^3-\cos x{)}^4$.
3. Differentiate the inner function with respect to $x$: $\frac{du}{dx}=6x^2+\sin x$ (the derivative of $-\cos x$ is $\sin x$, not $-\sin x$).
4. Multiply the two results per the chain rule, and simplify: $\frac{dy}{dx}=5(6x^2+\sin x)(2x^3-\cos x{)}^4$.

Exam tip: Always check for a nested inner function before applying a basic derivative rule—if the base/argument of your function is anything more complicated than just $x$, you will almost always need the chain rule.

3. Selecting Implicit Differentiation for Implicit Relations

An implicit relation is an equation relating $x$ and $y$ where $y$ cannot be easily solved explicitly as a function of $x$ (i.e., you cannot write it in the form $y=f(x)$ without very messy algebra, or at all). Common examples include $x^2+y^2=25$, $\sin (xy)=x+y$, and $e^{2y}+xy=0$. Implicit differentiation is the correct procedure for these cases. The method relies on the chain rule: since $y$ is a function of $x$, any term containing $y$ is a composite function of $x$, so you multiply by $\frac{dy}{dx}$ when you differentiate that term. The steps are: differentiate both sides of the equation with respect to $x$, collect all terms with $\frac{dy}{dx}$ on one side, factor out $\frac{dy}{dx}$, then solve for $\frac{dy}{dx}$.

Worked Example

Find $\frac{dy}{dx}$ for the relation $e^{2y}+x^{3}y=\sin x$.

1. Differentiate both sides term-by-term with respect to $x$: $\frac{d}{dx}[e^{2y}]+\frac{d}{dx}[x^{3}y]=\frac{d}{dx}[\sin x]$.
2. Apply the chain rule to the $e^{2y}$ term and product + chain rule to the $x^{3}y$ term: $2e^{2y}\frac{dy}{dx}+3x^{2}y+x^3\frac{dy}{dx}=\cos x$.
3. Collect all terms with $\frac{dy}{dx}$ on the left, all other terms on the right: $\frac{dy}{dx}\left(2e^{2y}+x^3\right)=\cos x-3x^{2}y$.
4. Divide both sides to solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{\cos x-3x^{2}y}{2e^{2y}+x^3}$.

Exam tip: If a question asks for the derivative at a specific point, substitute the point values into your derivative expression immediately after solving for $\frac{dy}{dx}$—you do not need to simplify further, which saves time.

4. Selecting the Inverse Function Derivative Rule

If $g(x)=f^{-1}(x)$ is the inverse function of $f(x)$, the inverse function derivative rule lets you find the derivative of $g(x)$ without having to solve for the inverse explicitly. The rule comes from implicit differentiation: if $y=f^{-1}(x)$, then $f(y)=x$. Differentiating both sides implicitly gives $f^{\prime }(y)\frac{dy}{dx}=1$, so rearranging gives the formula: $$\frac{d}{dx}\left[f^{-1}(x)\right]=\frac{1}{f^{\prime }\left(f^{-1}(x)\right)}$$ This rule is only valid when $f^{\prime }(f^{-1}(x))\ne 0$. This procedure is correct to use when you need the derivative of an inverse function at a point, and it also gives us the standard derivatives for inverse trigonometric functions (which are just inverse functions of basic trigonometric functions). When you have an inverse trigonometric function of a non-$x$ argument, you combine the inverse derivative rule with the chain rule.

Worked Example

Let $f(x)=x^3+2x+1$, and let $g(x)=f^{-1}(x)$. Find $g^{\prime }(4)$.

1. Apply the inverse derivative rule: $g^{\prime }(4)=\frac{1}{f^{\prime }(g(4))}$. First we need $g(4)=f^{-1}(4)$, which is the value of $x$ where $f(x)=4$.
2. Solve $f(x)=4$: $x^3+2x+1=4⟹x^3+2x-3=0$. Testing simple integer roots, $x=1$ satisfies the equation: $1+2-3=0$, so $g(4)=1$.
3. Compute $f^{\prime }(x)=3x^2+2$, so $f^{\prime }(g(4))=f^{\prime }(1)=3(1{)}^2+2=5$.
4. Substitute back into the inverse derivative formula: $g^{\prime }(4)=\frac{1}{5}$.

Exam tip: You never need to solve for the inverse explicitly to find its derivative at a point—AP problems are designed so the required $x$-value for the original function will always be a simple integer.

Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting the chain rule for the inner function, e.g., writing $\frac{d}{dx}\sin (3x)=\cos (3x)$ instead of $3\cos (3x)$. Why: Students only differentiate the outer function and stop, confusing composite functions with functions of $x$ directly. Correct move: Always ask "is the argument/base more than just $x$?" after differentiating the outer function, and multiply by the inner derivative if yes.
• Wrong move: Forgetting to multiply by $\frac{dy}{dx}$ when differentiating $y$-terms in implicit differentiation, e.g., writing $\frac{d}{dx}[y^2]=2y$ instead of $2y\frac{dy}{dx}$. Why: Students treat $y$ as a constant or independent variable, instead of a function of $x$. Correct move: Every time you differentiate a term containing $y$, immediately write $\frac{dy}{dx}$ after applying the derivative rule before moving to the next term.
• Wrong move: Reversing the inverse derivative formula, e.g., writing $(f^{-1}{)}^{\prime }(a)=f^{\prime }(f^{-1}(a))$ instead of $\frac{1}{f^{\prime }(f^{-1}(a))}$. Why: Students mix up the order when memorizing the formula. Correct move: If you forget the order, start from $f(y)=x$, differentiate implicitly, and rederive the formula in 10 seconds.
• Wrong move: Using implicit differentiation for an explicitly defined function, leading to overly complicated incorrect results, e.g., differentiating $y=\sin (x^2)$ implicitly instead of using chain rule. Why: Students default to implicit differentiation when it is not needed after learning the technique. Correct move: First check if you can easily solve for $y$ explicitly—only use implicit differentiation when you cannot.
• Wrong move: Forgetting the chain rule when differentiating inverse trig functions with non-$x$ arguments, e.g., writing $\frac{d}{dx}\arctan (2x)=\frac{1}{1+4x^2}$ instead of $\frac{2}{1+4x^2}$. Why: Students memorize the inverse trig derivative for $x$ and stop. Correct move: Treat any inverse trig function with a non-$x$ argument as a composite function, and multiply by the inner derivative.
• Wrong move: Trying to solve for an inverse explicitly to find its derivative at a point, leading to algebraic dead ends. Why: Students do not remember the inverse derivative rule works without an explicit inverse. Correct move: Always use the inverse derivative rule when asked for the derivative of an inverse at a point.

Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

What is $\frac{dy}{dx}$ for $y=\arcsin (3x^2)$? A) $\frac{1}{\sqrt{1-9x^4}}$ B) $\frac{6x}{\sqrt{1-9x^4}}$ C) $\frac{6x}{\sqrt{1-3x^2}}$ D) $\frac{1}{\sqrt{1-3x^2}}$

Worked Solution: This function is composite: the outer function is $\arcsin (u)$ and the inner function is $u=3x^2$. We use the inverse trig derivative rule for $\arcsin (u)$, which gives $\frac{d}{du}\arcsin (u)=\frac{1}{\sqrt{1-u^2}}$, then apply the chain rule by multiplying by $\frac{du}{dx}$. We find $\frac{du}{dx}=6x$ and $u^2=(3x^2{)}^2=9x^4$. Substituting gives $\frac{dy}{dx}=\frac{6x}{\sqrt{1-9x^4}}$. The correct answer is B.

Question 2 (Free Response)

Consider the curve defined by the relation $x^2+2xy+y^3=4$. (a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$ using implicit differentiation. (b) Find the slope of the tangent line to the curve at the point $(1,1)$. (c) Is the curve concave up or concave down at $(1,1)$? Justify your answer.

Worked Solution: (a) Differentiate both sides with respect to $x$: $$2x+2y+2x\frac{dy}{dx}+3y^2\frac{dy}{dx}=0$$ Collect terms with $\frac{dy}{dx}$ and factor: $$\frac{dy}{dx}(2x+3y^2)=-2x-2y⟹\frac{dy}{dx}=\frac{-2(x+y)}{2x+3y^2}$$

(b) Substitute $x=1,y=1$: $$\frac{dy}{dx}{|}_{(1,1)}=\frac{-2(1+1)}{2(1)+3(1{)}^2}=-\frac{4}{5}$$ The slope of the tangent line at $(1,1)$ is $-\frac{4}{5}$.

(c) To find concavity, compute the second derivative using the quotient rule. At $(1,1)$, the numerator of the second derivative simplifies to $-\frac{66}{5}$ (negative) and the denominator is $25$ (positive), so $\frac{d^{2}y}{d{x}^2}<0$. The curve is concave down at $(1,1)$, since the second derivative is negative at that point.

Question 3 (Application / Real-World Style)

The position of a particle moving along the $x$-axis at time $t$ (seconds) is given by $s(t)=\cos (e^{0.5t})$, where $s$ is measured in centimeters. Find the velocity of the particle at $t=2$ seconds, and include units. Round your answer to 3 decimal places.

Worked Solution: Velocity is the first derivative of position, so $v(t)=s^{\prime }(t)$. The position function is a double composite: outer $\cos (u)$, middle $e^v$, inner $v=0.5t$. Apply the chain rule twice: $$v(t)=-\sin (e^{0.5t})\cdot e^{0.5t}\cdot 0.5=-0.5e^{0.5t}\sin (e^{0.5t})$$ Substitute $t=2$, so $0.5t=1$: $e^1\approx 2.71828$, $\sin (e)\approx 0.41078$. Then: $$v(2)=-0.5(2.71828)(0.41078)\approx -0.559$$ The velocity of the particle at $t=2$ seconds is approximately $-0.559$ centimeters per second, meaning the particle is moving to the left at ~0.559 cm/s at that time.

Quick Reference Cheatsheet

What's Next

Mastering procedure selection for derivatives is the foundation for all remaining differentiation and integration topics in AP Calculus BC. Next, you will apply these techniques to parametric equations, polar curves, and vector-valued functions, where chain rule and implicit differentiation are used to find slopes of tangents and rates of change. Without the ability to quickly select the correct derivative procedure, these topics will be unnecessarily difficult, and you will waste valuable exam time second-guessing your choice of rule. Long-term, this skill is also critical for integration by substitution (where you reverse the chain rule) and for solving related rates and optimization problems. Follow-on topics to study next: Parametric and polar differentiation Integration by substitution Related rates Inverse trigonometric derivatives