Definition of the Derivative and Basic Rules: AP Calculus AB Math Worksheet Answer Key – PDF Download

AP Calculus Derivative Worksheet Answer Key: problems 1–5

Problem 1: Average rate of change

\(f(1)=4\) and \(f(4)=16+12=28\). The average rate is \(\frac{28-4}{4-1}=\frac{24}{3}=\) 8. It is the slope of the secant line through \((1,4)\) and \((4,28)\).

Problem 2: Derivative at a point by definition

1. \(\frac{(3+h)^{2}-9}{h}=\frac{6h+h^{2}}{h}=6+h\to\) 6
2. \(f(2)=12\). \(\frac{(2+h)^{2}+4(2+h)-12}{h}=\frac{4h+h^{2}+4h}{h}=8+h\to\) 8

Problem 3: Limit definition for a function

\(f(x+h)-f(x)=2(x^{2}+2xh+h^{2})-5x-5h-2x^{2}+5x=4xh+2h^{2}-5h\). Divide by \(h\): \(4x+2h-5\). Let \(h\to 0\): \(f^{\prime}(x)=4x-5\). Then \(f^{\prime}(1)=4-5=\) \(-1\).

Problem 4: Definition with a reciprocal

\(\frac{2}{x+h-1}-\frac{2}{x-1}=\frac{2(x-1)-2(x+h-1)}{(x-1)(x+h-1)}=\frac{-2h}{(x-1)(x+h-1)}\). Divide by \(h\): \(\frac{-2}{(x-1)(x+h-1)}\). Let \(h\to 0\): \(g^{\prime}(x)=-\frac{2}{(x-1)^{2}}\).

Problem 5: Power rule

1. \(8x^{7}\)
2. \(6\cdot 5x^{4}=\) \(30x^{4}\)
3. Write \(3x^{-2}\): \(-6x^{-3}=-\frac{6}{x^{3}}\)
4. Write \(x^{1/3}\): \(\frac{1}{3}x^{-2/3}\)

AP Calculus Derivative Worksheet Answer Key: problems 6–10

Problem 6: Sum and constant multiple rules

\(f^{\prime}(x)=\) \(12x^{2}-14x+5\). \(f^{\prime}(2)=48-28+5=\) 25. Solve \(12x^{2}-14x+5=5\): \(2x(6x-7)=0\), so \(x=0\) or \(x=\frac{7}{6}\).

Problem 7: Derivatives of sin, cos, e^x and ln x

1. \(3\cos x+2\sin x\)
2. \(5e^{x}+\frac{2}{x}\)
3. \(g^{\prime}(x)=e^{x}+\cos x\), so \(g^{\prime}(0)=1+1=\) 2
4. \(h^{\prime}(x)=\frac{4}{x}-3x^{2}\), so \(h^{\prime}(1)=4-3=\) 1

Problem 8: Tangent line

\(f(2)=8-4=4\). \(f^{\prime}(x)=3x^{2}-2\), so \(f^{\prime}(2)=10\). Point-slope form: \(y-4=10(x-2)\), so \(y=10x-16\).

Problem 9: Horizontal tangent lines

\(f^{\prime}(x)=6x^{2}-6x-12=6(x-2)(x+1)=0\), so \(x=2\) and \(x=-1\). \(f(2)=16-12-24+1=-19\) and \(f(-1)=-2-3+12+1=8\). The points are \((2,-19)\) and \((-1,8)\).

Problem 10: Product rule

1. \(2x(x^{3}-4x)+(x^{2}+1)(3x^{2}-4)=2x^{4}-8x^{2}+3x^{4}-x^{2}-4=\) \(5x^{4}-9x^{2}-4\). Check: expanding first gives \(x^{5}-3x^{3}-4x\), whose derivative is the same.
2. \(2x\sin x+x^{2}\cos x\)
3. \(e^{x}\ln x+\frac{e^{x}}{x}\)

AP Calculus Derivative Worksheet Answer Key: problems 11–15

Problem 11: Quotient rule

1. \(\frac{2x(x+1)-x^{2}\cdot 1}{(x+1)^{2}}=\) \(\frac{x^{2}+2x}{(x+1)^{2}}\)
2. \(\frac{x\cos x-\sin x}{x^{2}}\)
3. \(\frac{e^{x}x^{2}-e^{x}\cdot 2x}{x^{4}}=\) \(\frac{e^{x}(x-2)}{x^{3}}\)

Problem 12: Derivatives of tan, sec, csc and cot

1. \(f^{\prime}(x)=\text{sec}^{2}\,x\). At \(\frac{\pi}{4}\), \(\text{sec}\frac{\pi}{4}=\sqrt{2}\), so the value is \((\sqrt{2})^{2}=\) 2
2. \(3\,\text{sec}\,x\tan x\)
3. \(-2\,\text{csc}^{2}\,x-\text{csc}\,x\,\text{cot}\,x\)

Problem 13: Estimate derivatives from a table

1. \(\frac{5.0-2.0}{2-1}=\) 3
2. \(f^{\prime}(2)\approx\frac{7.7-3.1}{2.5-1.5}=\) 4.6
3. \(y-5=4.6(x-2)\), so \(y=4.6x-4.2\)

Problem 14: Is it differentiable?

1. Left of \(2\), \(f(x)=2-x\) with slope \(-1\); right of \(2\), \(f(x)=x-2\) with slope \(1\). The slopes differ, so not differentiable at \(x=2\) (a corner).
2. Continuity: \(1^{2}=1\) and \(2(1)-1=1\). Slopes: \(2x=2\) from the left and \(2\) from the right. They match, so \(p\) is differentiable and \(p^{\prime}(1)=2\).

Problem 15: Height of a ball

1. \(h(1)=37\) and \(h(2)=-64+96+5=37\), so the average velocity is 0 ft/s.
2. \(h^{\prime}(t)=-32t+48\), so \(h^{\prime}(1)=\) 16 ft/s (moving upward).
3. \(h^{\prime}(t)=0\) when \(t=1.5\). \(h(1.5)=-36+72+5=\) 41 ft at 1.5 s.

AP Calculus Derivative Worksheet Answer Key: problems 16–20

Problem 16: Marginal cost of bicycles

\(C^{\prime}(x)=0.04x+3\), so \(C^{\prime}(100)=\) 7. When 100 helmets are made, cost rises at about $7 per additional helmet. Exactly, \(C(101)-C(100)=0.02(201)+3=\) $7.02, very close to the derivative.

Problem 17: Learning curve

\(S^{\prime}(t)=\frac{15}{t}\). \(S^{\prime}(5)=\) 3 words per minute per hour and \(S^{\prime}(10)=\) 1.5. Each extra hour of practice helps less as time goes on.

Problem 18: Revenue growth

\(p(10)=25\), \(q(10)=260\), \(p^{\prime}=0.5\), \(q^{\prime}=-4\). Product rule: \(R^{\prime}(10)=0.5(260)+25(-4)=130-100=\) 30. Revenue is growing by about $30 per day on day 10.

Problem 19: Concentration of a medicine

1. \(\frac{1(t^{2}+4)-t(2t)}{(t^{2}+4)^{2}}=\) \(\frac{4-t^{2}}{(t^{2}+4)^{2}}\)
2. \(C^{\prime}(1)=\frac{3}{25}=\) 0.12. At 1 hour the concentration is rising by about 0.12 mg/L per hour.
3. \(C^{\prime}(t)=0\) at \(t=2\) (for \(t > 0\)); \(C^{\prime} > 0\) before and \(C^{\prime} < 0\) after. \(C(2)=\frac{2}{8}=\) 0.25 mg/L at 2 hours.

Problem 20: Derivatives from a table of values

1. \(h^{\prime}(2)=f^{\prime}g+fg^{\prime}=(-1)(5)+(3)(4)=\) 7
2. \(k^{\prime}(2)=\frac{f^{\prime}g-fg^{\prime}}{g^{2}}=\frac{-5-12}{25}=\) \(-\frac{17}{25}\)
3. \(m^{\prime}(2)=2(-1)-3(4)=\) \(-14\)

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