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Module 2 Practice Problems: The Derivative

Instructions: Work through each problem, then reveal the solution to check your work.

Problem 1

Find the average rate of change of f(x) = x³ − 2x from x = 1 to x = 3.

Solution

[f(3) − f(1)]/(3−1) = [(27−6) − (1−2)]/2 = [21 − (−1)]/2 = 22/2 = 11.

Problem 2

Use the limit definition to find f'(x) for f(x) = 2x² + 3.

Solution

[2(x+h)²+3 − (2x²+3)]/h = [4xh+2h²]/h = 4x+2h. Limit: 4x.

Problem 3

Find f'(x) for f(x) = 5x4 − 3x² + 7x − 1.

Solution

20x³ − 6x + 7.

Problem 4

Find g'(t) for g(t) = 3/t + 2√t.

Solution

Rewrite: 3t−1 + 2t1/2. g'(t) = −3t−2 + t−1/2 = −3/t² + 1/√t.

Problem 5

Find the equation of the tangent line to y = x² − 3x at x = 2.

Solution

y(2) = 4−6 = −2. y' = 2x−3, y'(2) = 1. Tangent: y+2 = 1(x−2), y = x − 4.

Problem 6

Find where the tangent to f(x) = x³ − 3x is horizontal.

Solution

f'(x) = 3x² − 3 = 0. x² = 1, so x = ±1.

Problem 7

Find f''(x) for f(x) = x5 − 4x³ + x.

Solution

f' = 5x4 − 12x² + 1. f'' = 20x³ − 24x.

Problem 8

Find dy/dx for y = 4ex − 2 ln x + x3.

Solution

4ex − 2/x + 3x².

Problem 9

Is f(x) = |2x − 6| differentiable at x = 3?

Solution

No. f has a corner at x = 3. Left derivative: −2. Right derivative: 2. They differ.

Problem 10

A particle moves with position s(t) = t³ − 6t² + 9t. Find its velocity and acceleration at t = 2.

Solution

v(t) = s'(t) = 3t² − 12t + 9. v(2) = 12 − 24 + 9 = −3.

a(t) = v'(t) = 6t − 12. a(2) = 12 − 12 = 0.

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