Selecting procedures for calculating derivatives — AP Calculus AB
1. What Is Selecting Differentiation Procedures? ★★☆☆☆ ⏱ 3 min
Selecting procedures for calculating derivatives is the core skill of choosing the correct differentiation technique based on the algebraic structure of the function you need to differentiate, rather than just applying a pre-specified rule. Per the AP Calculus AB CED, this topic is part of Unit 3 and makes up 9–13% of the total AP exam score, tested in both multiple-choice and free-response sections. It is foundational for nearly every subsequent AP Calculus AB topic from related rates to curve sketching.
2. Recognizing and Applying the Chain Rule for Composite Functions ★★☆☆☆ ⏱ 4 min
You should always select the chain rule when working with a composite function. For nested composites with more than two layers, repeat the chain rule process working from the outermost layer inward. The chain rule formula is:
\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)
In Leibniz notation, if $y = f(u)$ and $u = g(x)$ this becomes:
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
Exam tip: Always work from the outermost layer inward when applying the chain rule; don't try to differentiate the inner layer first, which leads to missing a multiplication step.
3. Selecting Implicit Differentiation for Non-Explicit Functions ★★★☆☆ ⏱ 3 min
Implicit differentiation is the procedure to select when $y$ is not explicitly isolated as a function of $x$, meaning you cannot easily rewrite the relationship as $y = f(x)$. The core of implicit differentiation is just an application of the chain rule: any term containing $y$ is a composite function, so you must multiply by $\frac{dy}{dx}$ when differentiating it. The step-by-step procedure is:
- Differentiate every term on both sides of the equation with respect to $x$
- Apply the chain rule to all terms containing $y$, adding a $\frac{dy}{dx}$ factor
- Collect all terms with $\frac{dy}{dx}$ on one side of the equation
- Factor out $\frac{dy}{dx}$ and divide to solve for it
Exam tip: When finding $\frac{dy}{dx}$ at a specific point, you can substitute the point values into the equation before solving for $\frac{dy}{dx}$ to simplify algebra, which reduces the chance of arithmetic errors.
4. Selecting the Inverse Function Derivative Rule ★★★☆☆ ⏱ 3 min
You should select the inverse function derivative rule when you need the derivative of an inverse function at a point, and you do not want to (or cannot) find the inverse function explicitly. This is the most common AP exam question type for inverse derivatives, testing your ability to apply the rule rather than just do routine differentiation.
The intuition for this rule comes from the fact that the graph of an inverse function is the reflection of the original function over the line $y=x$, so the slope of the tangent of the inverse is the reciprocal of the slope of the original function at the corresponding point. To apply the rule: 1) find $b$ such that $f(b) = a$ (so $f^{-1}(a) = b$), 2) find $f'(b)$, 3) take the reciprocal.
Exam tip: If you are asked for $\left(f^{-1}\right)'(a)$, always solve $f(b) = a$ first before taking the derivative. You will almost always get an integer for $b$ on the AP exam, so testing small integers first saves time.
5. Combining Procedures for Complex Functions ★★★★☆ ⏱ 4 min
Most AP exam problems require combining multiple differentiation procedures. A common pattern is combining product/quotient rule with chain rule for factors that are composite functions. Always identify the top-level structure first: if the function is a product of two terms, apply product rule first, then use chain rule for any composite factors.
Common Pitfalls
Why: Students only differentiate the outer layer and stop, confusing composite functions with power functions of $x$.
Why: Students treat $y$ as a constant instead of a function of $x$ when differentiating.
Why: Confusion between the input to $f'$ and the input for the inverse function.
Why: Confusing a product of two functions with a composition of two functions.
Why: Mixing up the dependent and independent variable.