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Quizzes > Mathematics

AP Calculus unit 2 test: Differentiation MCQ Practice (AB)

Quick, free calculus unit 2 practice test with instant scoring and explanations.

Editorial: Review CompletedCreated By: Gabriel AzevedoUpdated Aug 28, 2025
Difficulty: Moderate
Grade: Grade 12
Study OutcomesCheat Sheet
Colorful paper art promoting AP Calculus AB Unit 2 Blitz trivia quiz for high school students.

Use this AP Calculus AB Unit 2 differentiation quiz to check your grasp of derivatives, tangent lines, and rates of change. It's a fast set with instant feedback to build speed and find gaps before the exam. For a quick refresh, try the AP calculus unit 1 test or focus on rules with the derivative rules quiz, then round it out with a broader calculus quiz.

Using the limit definition, the derivative of f(x) = x^2 at x = 3 equals
6 - Explanation: f′(a)=lim_{h->0}[(a+h)^2-a^2]/h=lim(2a+h)=2a=6 for a=3.
3
12
9
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Apply the power rule to find d/dx of x^7.
7x^6 - Explanation: d/dx[x^n]=n x^(n-1).
x^6
6x^7
7x^7
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Find d/dx of 7sin(x).
7sin(x)
cos(x)
-7cos(x)
7cos(x) - Explanation: constant multiple rule with d/dx[sin x]=cos x.
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Compute d/dx of 3x^2 - 5x + 4.
6x^2 - 5
6x - 5 - Explanation: derivative is termwise, constants vanish.
6x - 5x
3x^2 - 5
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Compute d/dx of sin(x^2).
cos(x)
2x cos(x^2) - Explanation: chain rule with outer sin, inner x^2.
2x sin(x^2)
cos(x^2)
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Find d/dx of e^{3x}.
3e^{3x} - Explanation: chain rule on e^{u} gives u′e^{u} with u=3x.
e^x
3e^x
e^{3x}
undefined
Find d/dx of f(x)=x^2 e^x using product rule.
x^2 e^x + 2x e^x - Explanation: (uv)′=u′v+uv′ gives 2x e^x + x^2 e^x.
x^2 e^x
(2x)e^x - x^2 e^x
2x e^x
undefined
Differentiate g(x) = (ln x)/x for x>0.
(1 - ln x)/x^2 - Explanation: quotient rule; g′=(x*(1/x)-ln x*1)/x^2 = (1- ln x)/x^2.
ln x
(1 + ln x)/x^2
1/(x ln x)
undefined
For the circle x^2 + y^2 = 25, find dy/dx at the point (3,4).
3/4
4/3
-4/3
-3/4 - Explanation: 2x + 2y y′=0 so y′= -x/y = -3/4 at (3,4).
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Find the equation of the normal line to f(x)=√x at x=4.
y - 2 = -4(x - 4) - Explanation: f(4)=2, f′(x)=1/(2√x), f′(4)=1/4 so normal slope=-4.
y - 2 = 4(x - 4)
y - 4 = -4(x - 2)
y - 2 = -1/4(x - 4)
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Which hypothesis is NOT required for Rolle's Theorem on [a,b]?
f is concave on [a,b] - Explanation: Rolle requires continuity on [a,b], differentiability on (a,b), and f(a)=f(b).
f is continuous on [a,b]
f is differentiable on (a,b)
f(a) = f(b)
undefined
Given a differentiable f on [1,5] with f'(x)=3 for all x, what does the Mean Value Theorem guarantee?
f(5)=f(1)
There exists c in (1,5) with f'(c)= [f(5)-f(1)]/(5-1) - Explanation: MVT equates some f'(c) to average rate.
f is concave up on (1,5)
f'(x)=0 for some x in (1,5)
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Implicitly differentiate y^3 + x y = 7 to find dy/dx.
y′ = -x/(3y^2 - y)
y′ = -(y + x y′)/ (3y^2) leads to y′ = -y/(3y^2 - x) - Explanation: Differentiate: 3y^2 y′ + y + x y′ = 0, solve for y′.
y′ = (y - x)/(3y^2)
y′ = -y/(3y^2 + x)
undefined
A 10 ft ladder leans against a wall. The bottom slides away at 3 ft/s. When the bottom is 6 ft from the wall, how fast is the top sliding down?
-2.25 ft/s - Explanation: x^2+y^2=100, 2x x′+2y y′=0 at x=6 gives y=8, so y′= -(x/y) x′=-(6/8)*3= -2.25.
2.25 ft/s
-3.75 ft/s
-1.8 ft/s
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If g = f^{-1} and f is differentiable with f′(g(a)) ≠ 0, then g′(a) equals
1 / f′(g(a)) - Explanation: derivative of inverse function.
f(g(a))
1 / f′(a)
f′(a)
undefined
Differentiate y = x^x for x>0.
y′ = x^x
y′ = x^{x-1}
y′ = x^x(ln x + 1) - Explanation: log differentiate ln y = x ln x, then y′/y = ln x + 1.
y′ = x^x ln x
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For piecewise f(x)= { ax^2+bx, x<1 ; cx+d, x≥1 }, differentiable at x=1 implies
a=b=c=d
Only 2a + b = c
Only a+b = c+d
Continuity and equal derivatives: a+b = c+d and 2a + b = c - Explanation: match function and slope at the join.
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For x^2 y + y = 1, find dy/dx.
y′ = -2x/(x^2 + 1)
y′ = (2xy + 1)/(x^2 + 1)
y′ = -(2xy + 1)/(x^2 + 1) - Explanation: differentiate: 2x y + x^2 y′ + y′ = 0, group y′(x^2+1)=-(2xy+1).
y′ = -(x^2 + 1)/(2xy + 1)
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For x^2 + y^2 = 1, find d^2y/dx^2 in terms of x and y.
d^2y/dx^2 = -1/y^3 - Explanation: y′= -x/y; differentiate: y′′= -(y - x y′)/y^2 = -(y + x(x/y))/y^2=-(y^2+x^2)/(y^3)= -1/y^3.
d^2y/dx^2 = -1/y
d^2y/dx^2 = 1/y^3
d^2y/dx^2 = -x/y^2
undefined
Implicitly differentiate sin(xy) = x to find dy/dx.
y′ = [1 + y cos(xy)]/[x cos(xy)]
y′ = [1 - x cos(xy)]/[y cos(xy)]
y′ = [1 - y cos(xy)]/[x cos(xy)] - Explanation: cos(xy)(x y′ + y) = 1, solve for y′.
y′ = [y cos(xy) - 1]/[x cos(xy)]
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0

Study Outcomes

  1. Understand the definition of limits to determine function continuity.
  2. Apply derivative rules to compute instantaneous rates of change.
  3. Analyze function behavior to identify points of non-differentiability.
  4. Synthesize derivative concepts to solve real-world optimization problems.

AP Calc Unit 2 MCQ Test Cheat Sheet

  1. Definition of a Derivative - Think of the derivative as the ultimate speedometer for any function: it measures the instantaneous rate of change at a single point by taking the limit of the average rate of change. Mastering this concept sets the stage for everything else in calculus - no pressure, but it's kind of a big deal!
  2. Power Rule - This rule is your secret weapon for differentiating powers of x: if f(x)=x❿, then f′(x)=n·x❿❻¹. With this in your toolkit, polynomials go from scary to snack-sized in no time.
  3. Product Rule - When two differentiable functions u(x) and v(x) decide to team up, the derivative is u′v + uv′. It's a little like baking a cake: you mix and match ingredients (derivatives) to get the final result.
  4. Quotient Rule - Dividing functions? Use (u/v)′ = (u′v - uv′)/v² to keep things neat and tidy. Just remember to "low d-high, high d-low, square the bottom" - it's the chorus you never knew you needed!
  5. Chain Rule - For composite functions y=f(g(x)), the derivative is f′(g(x))·g′(x). It's like peeling an onion: differentiate the outer layer, then the inner one, and voilà, you've got your derivative.
  6. Differentiability Implies Continuity - If a function is differentiable at a point, it must be continuous there - but a continuous function isn't always differentiable. This subtle distinction is key for spotting tricky exceptions.
  7. Non-Differentiable Points - Watch out for corners, cusps, vertical tangents, and jumps - these are the culprits that break differentiability. Spotting them is like being a detective on a function's graph!
  8. Geometric Interpretation - The derivative at a point is the slope of the tangent line to the curve - imagine zooming in so close that the curve looks like a straight line. It's calculus meets geometry in the coolest way.
  9. Function Behavior Analysis - Use derivatives to find where functions rise, fall, and hit their peaks or valleys. It's like reading the mood swings of your graph!
  10. Trig Function Derivatives - Get sin, cos, and tan under your thumb with their specific rules: (sin x)′=cos x, (cos x)′= - sin x, and (tan x)′=sec² x. These are must-know moves for any calculus showdown.
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