← Back to the Derivative Rules worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Each solution shows the rule used and the algebra after it. If your result differs, check the sign of each term, the order of terms in the quotient rule, and the units in context problems.
Standards: AP Calculus AB Unit 2 · Level: Grade 12 AP Calculus AB
Problem 1 Average rate of change
f(4) = 64 − 20 = 44 and f(1) = 1 − 5 = −4. The average rate is 44 − (−4)4 − 1 = 48/3 = 16.
Problem 2 Rate of change from a table
1,630 − 1,3106 − 2 = 320/4 = 80 feet per mile.
Problem 3 Power rule with a polynomial
Differentiate each term: 7·4x3 − 5·2x + 0. So f′(x) = 28x3 − 10x.
Problem 4 Power rule with roots and fractions
Rewrite as 10x1/2 − 12x−4. Then f′(x) = 5x−1/2 + 48x−5 = 5√x + 48x5.
Problem 5 Derivatives of sine and cosine
The derivative of sin x is cos x, and the derivative of cos x is −sin x. So f′(x) = 5 cos x − 2(−sin x) = 5 cos x + 2 sin x.
Problem 6 Exponential and logarithm terms
Differentiate term by term: 3ex + 4/x − 1. So f′(x) = 3ex + 4x − 1.
Problem 7 Limit definition at a point
f(5) = 25 + 35 = 60. Then (5 + h)2 + 7(5 + h) − 60h = 17h + h2h = 17 + h. As h → 0 the limit is 17.
Problem 8 Product rule
Product rule: 6x(x3 − 4x) + (3x2 + 1)(3x2 − 4). Expand: 6x4 − 24x2 + 9x4 − 12x2 + 3x2 − 4. Combine: f′(x) = 15x4 − 33x2 − 4.
Problem 9 Quotient rule
f′(x) = 2(x2 + 1) − (2x + 5)(2x)(x2 + 1)2. The numerator is 2x2 + 2 − 4x2 − 10x = −2x2 − 10x + 2. So f′(x) = −2x2 − 10x + 2(x2 + 1)2.
Problem 10 Equation of a tangent line
f(2) = 8 − 6 + 1 = 3. Since f′(x) = 3x2 − 3, f′(2) = 12 − 3 = 9. Point-slope form: y − 3 = 9(x − 2), so y = 9x − 15.
Problem 11 Derivatives of tangent and secant
The derivative of tan x is sec2 x and the derivative of sec x is sec x tan x. So f′(x) = sec2 x − 2 sec x tan x.
Problem 12 Product rule with a trig factor
f′(x) = 2x sin x + x2 cos x. At x = π: 2π(0) + π2(−1) = −π2.
Problem 13 Quotient rule with a logarithm
f′(x) = (1/x)x2 − (ln x)(2x)x4 = x − 2x ln xx4 = 1 − 2 ln xx3. At x = e, ln e = 1, so f′(e) = −1e3.
Problem 14 Differentiable piecewise function
Continuity: the left value is 9, so 3a + b = 9. Equal slopes: the left derivative is 2x = 6, so a = 6. Then b = 9 − 18 = −9. The values are a = 6 and b = −9.
Problem 15 Checking differentiability at a corner
For x < 3, f(x) = 3 − x, with slope −1. For x > 3, f(x) = x − 3, with slope +1. The one-sided derivatives disagree, so f′(3) does not exist even though f is continuous at 3.
Problem 16 Estimating a derivative from a table
Use the values on either side of 4: D(6) − D(2)6 − 2 = (44 − 18)/4 = 26/4 = 6.5 feet per hour. The river is rising at about 6.5 ft/h at t = 4.
Problem 17 Horizontal tangent lines
Set f′(x) = 3x2 − 12 = 0, so x2 = 4 and x = ±2. Then f(2) = 8 − 24 + 5 = −11 and f(−2) = −8 + 24 + 5 = 21. The points are (2, −11) and (−2, 21).
Problem 18 Drone height
h(1) = 32 and h(3) = 72, so the average velocity is 40/2 = 20 ft/s. The derivative is h′(t) = −8t + 36, so h′(2) = 20 ft/s. Setting −8t + 36 = 0 gives t = 4.5 s, when the drone reaches its top height h(4.5) = 81 ft.
Problem 19 Concentration of a medicine
Quotient rule: C′(t) = 8(t2 + 4) − 8t(2t)(t2 + 4)2 = 32 − 8t2(t2 + 4)2. Then C′(1) = 24/25 = 0.96 mg/L per hour. The rate is zero when 32 = 8t2, so t = 2 hours, when C(2) = 2 mg/L.
Problem 20 Tangent line through the origin
Let the point of tangency be (a, ln a) with slope 1/a. The tangent line is y − ln a = (1/a)(x − a). Substituting (0, 0): −ln a = −1, so ln a = 1 and a = e. The slope is 1/e. The point is (e, 1) and the line is y = x/e.
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