AP Calculus AB › Unit 3
Unit 3 is about the chain rule and everything built on it. You will differentiate composite functions, use implicit differentiation, and find derivatives of inverse functions, including inverse trig functions. Higher-order derivatives also appear. Exam questions often give values of functions and their derivatives in a table and ask for the derivative of a composition. The most common error is forgetting to multiply by the derivative of the inside function. In implicit differentiation, students also forget that y depends on x, so every y term needs a dy/dx.
These are taken from the Regular set, with the answer and explanation shown.
1. A differentiable function f has f(2) = π and f′(2) = 3. If h(x) = sin(f(x)), what is h′(2)?
Answer: D. h′(x) = cos(f(x))·f′(x). So h′(2) = cos(π)·3 = (−1)(3) = −3.
Common trap: Choosing 0 computes sin(π), the value of h(2). The derivative uses cos(π), not sin(π).
2. Let f(x) = 5 − x³, and let g be the inverse of f. What is g′(−3)?
Answer: C. Solve 5 − x³ = −3: x³ = 8, so g(−3) = 2. Then g′(−3) = 1/f′(2) = 1/(−12) = −1/12.
Common trap: Choosing −1/27 evaluates f′ at −3. The value −3 is an output; the matching input is 2.
3. Let f(x) = x sin(x²). What is the least positive value of x at which f′(x) = 0? Round to three decimal places.
Answer: B. By the product and chain rules, f′(x) = sin(x²) + 2x²cos(x²). Solving f′(x) = 0 on a calculator gives the least positive solution x ≈ 1.355.
Common trap: Choosing 1.772 gives √π, where f(x) itself is 0. A zero of f is not the same as a zero of f′.