Source: AP Practice Questions, Units 2 & 3
Tags: tangent lines, implicit differentiation, inverse functions, differentiability, continuity, AP Calculus AB
This set covers how derivatives are used rather than just computed: finding equations of tangent lines, differentiating implicitly when y cannot be isolated, working with inverse function derivatives, and determining when a piecewise function is differentiable. These are the application-layer questions that appear throughout the AP exam's multiple-choice and free-response sections.
Tangent line
A line that touches a curve at exactly one point (locally) and has a slope equal to the derivative at that point. Equation form: y - y₁ = m(x - x₁), where m = f'(x₁).
Implicit differentiation
A technique for finding dy/dx when y is not isolated. Differentiate both sides with respect to x, treating y as a function of x (so every time you differentiate a y-term, multiply by dy/dx), then solve for dy/dx.
Inverse function derivative
If g(x) = f⁻¹(x), then g'(x) = 1 / f'(g(x)). The derivative of the inverse at a point is the reciprocal of the original function's derivative, evaluated at the corresponding point.
Differentiable
A function is differentiable at a point if it has a derivative there. This requires continuity, plus the left-hand and right-hand derivatives must agree.
The exam tests tangent lines in several forms: find the equation, find the slope, or use the tangent line to determine a constant.
Finding the tangent line directly:
If f(x) = 4x³ - 5x + 3, find the tangent line at x = -1.
f(-1) = 4(-1) - 5(-1) + 3 = -4 + 5 + 3 = 4
f'(x) = 12x² - 5, so f'(-1) = 12 - 5 = 7
Tangent: y - 4 = 7(x + 1), which gives y = 7x + 11
Answer: (C)
Using a tangent line that passes through a given point:
The tangent to f at (1, 7) passes through (-2, -2). Find f'(1).
Slope = (7 - (-2)) / (1 - (-2)) = 9/3 = 3
So f'(1) = 3
Answer: (C)
Tangent line matching a given line:
The line x + y = k is tangent to y = x² + 3x + 1. Find k.
The line has slope -1. Set y' = 2x + 3 = -1, giving x = -2.
At x = -2: y = 4 - 6 + 1 = -1.
k = x + y = -2 + (-1) = -3.
Answer: (A)
Find the slope of the tangent to y = arctan(4x) at x = 1/4.
y' = 4 / (1 + 16x²)
At x = 1/4: y' = 4 / (1 + 16 · 1/16) = 4 / 2 = 2
Answer: (A)
When an equation mixes x and y and you cannot solve for y, differentiate both sides with respect to x. Every term involving y picks up a dy/dx factor via the chain rule.
If sin(xy) = x, find dy/dx.
Differentiate: cos(xy) · (y + x · dy/dx) = 1
Expand: y cos(xy) + x cos(xy) · dy/dx = 1
Isolate: dy/dx = [1 - y cos(xy)] / [x cos(xy)]
Answer: (D)
The key step people miss: when differentiating xy inside sin(xy), you need the product rule on xy, giving y + x · dy/dx. Both terms matter.
A piecewise function is differentiable at the join point only if two conditions hold: continuity (the pieces agree at the boundary) and equal derivatives from both sides.
f(x) = cx + d for x ≤ 2, and f(x) = x² - cx for x > 2. Find c + d.
Continuity at x = 2: 2c + d = 4 - 2c, so d = 4 - 4c.
Equal derivatives at x = 2: Left derivative is c. Right derivative is 2x - c = 4 - c at x = 2. Setting c = 4 - c gives c = 2.
Then d = 4 - 8 = -4, so c + d = -2.
Answer: (B)
If g = f⁻¹, you need to find the right corresponding point. Since g(x) = f⁻¹(x), the value g(a) is the x-value where f equals a.
Given f(3) = 15, f(6) = 3, f'(3) = -8, f'(6) = -2, and g(x) = f⁻¹(x), find g'(3).
g(3) = f⁻¹(3). Since f(6) = 3, we have g(3) = 6.
g'(3) = 1 / f'(g(3)) = 1 / f'(6) = 1 / (-2) = -1/2.
Answer: (A)
The formula g'(x) = 1/f'(g(x)) is worth memorising. The common mistake is evaluating f' at the wrong point.
Given: y³ - xy = 2, with dy/dx = y / (3y² - x).
(a) Tangent line at (-1, 1):
dy/dx at (-1, 1) = 1 / (3(1) - (-1)) = 1/4
Tangent: y - 1 = (1/4)(x + 1), or y = (1/4)x + 5/4
(b) Vertical tangent points:
Vertical tangent occurs where the denominator 3y² - x = 0, so x = 3y².
Substitute into y³ - xy = 2: y³ - 3y³ = 2, giving -2y³ = 2, so y = -1.
Then x = 3(1) = 3. The point is (3, -1).
(c) Second derivative at (-1, 1):
d²y/dx² requires differentiating dy/dx = y/(3y² - x) using the quotient rule.
Numerator of quotient rule: (dy/dx)(3y² - x) - y(6y · dy/dx - 1)
With dy/dx = 1/4 at (-1, 1):
= (1/4)(3 + 1) - 1(6 · 1/4 - 1) = 1 - 1/2 = 1/2
Denominator: (3y² - x)² = (4)² = 16
d²y/dx² = (1/2)/16 = 1/32
Given: f(0) = 2, f'(0) = -4, f''(0) = 3.
(a) g(x) = e^(ax) + f(x). Find g'(0) and g''(0) in terms of a.
g'(x) = ae^(ax) + f'(x), so g'(0) = a + (-4) = a - 4
g''(x) = a²e^(ax) + f''(x), so g''(0) = a² + 3
(b) h(x) = cos(kx) · f(x). Find h'(x) and the tangent line at x = 0.
Product rule with chain rule: h'(x) = -k sin(kx) · f(x) + cos(kx) · f'(x)
h'(0) = -k · 0 · 2 + 1 · (-4) = -4
h(0) = cos(0) · f(0) = 1 · 2 = 2
Tangent line: y = -4x + 2
Note: the constant k vanishes at x = 0 because sin(0) = 0. This is a deliberate feature of the question, testing whether you carry through the algebra rather than guessing that k must appear in the answer.
Tangent line equation:
y - f(a) = f'(a)(x - a)
Inverse function derivative:
If g = f⁻¹, then g'(x) = 1 / f'(g(x))
Implicit differentiation process:
Differentiate every term with respect to x.
Apply chain rule to y-terms (multiply by dy/dx).
Collect all dy/dx terms on one side.
Factor out dy/dx and solve.
Differentiability checklist (piecewise):
Continuity: left-hand value = right-hand value at the join.
Derivatives match: left-hand derivative = right-hand derivative at the join.
Inverse trig derivatives:
d/dx[arctan(u)] = u' / (1 + u²)
d/dx[arcsin(u)] = u' / √(1 - u²)
⚠️ For inverse function derivatives, the most common error is plugging into f' at the wrong input. You need f' evaluated at g(x), not at x itself.
⚠️ Implicit differentiation requires the product rule whenever x and y are multiplied together. Do not forget the dy/dx factor on every y-term.
⚠️ Piecewise differentiability questions always require two equations: one for continuity, one for matching derivatives. Solving only one of these is the classic half-credit mistake on FRQs.
⚠️ When a tangent line "passes through" a point, that point may not be on the curve. Use the two-point slope formula, not the derivative at the external point.
⚠️ Vertical tangent lines occur where the denominator of dy/dx equals zero (and the numerator does not). This is a common FRQ setup.
Q: The line x + y = k is tangent to y = x² + 3x + 1. What is k?
A: -3. The line has slope -1, so set 2x + 3 = -1 to find x = -2, then y = -1, and k = -2 + (-1) = -3.
Q: If sin(xy) = x, what is dy/dx?
A: (1 - y cos(xy)) / (x cos(xy)). Differentiate implicitly, apply the product rule to xy inside the cosine, then isolate dy/dx.
Q: For f(x) = {cx + d, x ≤ 2; x² - cx, x > 2}, if f is differentiable at x = 2, what is c + d?
A: -2. Continuity gives d = 4 - 4c. Matching derivatives gives c = 2. Then d = -4.
Q: If g(x) = f⁻¹(x), f(6) = 3, and f'(6) = -2, what is g'(3)?
A: -1/2. Since f(6) = 3, g(3) = 6. Then g'(3) = 1/f'(6) = -1/2.
Q: The tangent to f at (1, 7) passes through (-2, -2). What is f'(1)?
A: 3. The slope between the two points is (7 - (-2))/(1 - (-2)) = 9/3 = 3.
Q: Find the slope of y = arctan(4x) at x = 1/4.
A: 2. The derivative is 4/(1 + 16x²). At x = 1/4 this gives 4/2 = 2.
tangent line equation, point-slope form, implicit differentiation, dy/dx, inverse function derivative, piecewise differentiability, continuity, arctan derivative, inverse trig, vertical tangent, second derivative implicit, AP Calculus AB, Unit 2, Unit 3, free response, 2015 AP exam