Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

  1. My Subjects
  3. AP Calculus BC

  5. Flashcards

AP Calculus BC Flashcards: Estimating Derivatives Of A Function

Study Estimating Derivatives Of A Function in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

What this deck covers

This deck focuses on Estimating Derivatives Of A Function, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Estimating Derivatives Of A Function

1

/ 30

0 reviewed

0% Complete

0 reviewing
QUESTION

What is the formula for the backward difference quotient to estimate the derivative at a point?

Tap or drag to reveal answer

ANSWER

f′(a)≈f(a)−f(a−h)hf'(a) \approx \frac{f(a) - f(a-h)}{h}f′(a)≈hf(a)−f(a−h)​. Uses the point before aaa to estimate the slope.

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: What is the formula for the backward difference quotient to estimate the derivative at a point?

Answer: f′(a)≈f(a)−f(a−h)hf'(a) \approx \frac{f(a) - f(a-h)}{h}f′(a)≈hf(a)−f(a−h)​. Uses the point before aaa to estimate the slope.

Flashcard 2: Estimate f′(3)f'(3)f′(3) for f(x)=tan⁡−1(x)f(x) = \tan^{-1}(x)f(x)=tan−1(x) using h=0.01h = 0.01h=0.01 and the backward difference quotient.

Answer: 0.0995. Backward formula: tan⁡−1(3)−tan⁡−1(2.99)0.01≈0.0995\frac{\tan^{-1}(3) - \tan^{-1}(2.99)}{0.01} \approx 0.09950.01tan−1(3)−tan−1(2.99)​≈0.0995

Flashcard 3: Estimate f′(1)f'(1)f′(1) for f(x)=2xf(x) = 2^xf(x)=2x using h=0.001h = 0.001h=0.001 and the symmetric difference quotient.

Answer: 1.3863. Symmetric formula: 21.001−20.9990.002≈1.3863\frac{2^{1.001} - 2^{0.999}}{0.002} \approx 1.38630.00221.001−20.999​≈1.3863

Flashcard 4: Estimate f′(0)f'(0)f′(0) for f(x)=log⁡(1+x)f(x) = \log(1+x)f(x)=log(1+x) using h=0.01h = 0.01h=0.01 and the symmetric difference quotient.

Answer: 0.9950. Symmetric formula: log⁡(1.01)−log⁡(0.99)0.02≈0.9950\frac{\log(1.01) - \log(0.99)}{0.02} \approx 0.99500.02log(1.01)−log(0.99)​≈0.9950

Flashcard 5: Estimate f′(4)f'(4)f′(4) for f(x)=1x3f(x) = \frac{1}{x^3}f(x)=x31​ using h=0.1h = 0.1h=0.1 and the forward difference quotient.

Answer: -0.0351. Forward formula: 1/(4.1)3−1/640.1≈−0.0351\frac{1/(4.1)^3 - 1/64}{0.1} \approx -0.03510.11/(4.1)3−1/64​≈−0.0351

Flashcard 6: Estimate f′(1)f'(1)f′(1) for f(x)=sin⁡(2x)f(x) = \sin(2x)f(x)=sin(2x) using h=0.001h = 0.001h=0.001 and the symmetric difference quotient.

Answer: 1.0806. Symmetric formula: sin⁡(2.002)−sin⁡(1.998)0.002≈1.0806\frac{\sin(2.002) - \sin(1.998)}{0.002} \approx 1.08060.002sin(2.002)−sin(1.998)​≈1.0806

Flashcard 7: Estimate f′(2)f'(2)f′(2) for f(x)=tan⁡xf(x) = \tan xf(x)=tanx using h=0.01h = 0.01h=0.01 and the forward difference quotient.

Answer: 2.0199. Forward formula: tan⁡(2.01)−tan⁡(2)0.01≈2.0199\frac{\tan(2.01) - \tan(2)}{0.01} \approx 2.01990.01tan(2.01)−tan(2)​≈2.0199

Flashcard 8: Estimate f′(1)f'(1)f′(1) for f(x)=xf(x) = \sqrt{x}f(x)=x​ using h=0.001h = 0.001h=0.001 and the backward difference quotient.

Answer: 0.4995. Backward formula: 1−0.9990.001≈0.4995\frac{\sqrt{1} - \sqrt{0.999}}{0.001} \approx 0.49950.0011​−0.999​​≈0.4995

Flashcard 9: Estimate f′(5)f'(5)f′(5) for f(x)=log⁡xf(x) = \log xf(x)=logx using h=0.01h = 0.01h=0.01 and the backward difference quotient.

Answer: 0.1990. Backward formula: log⁡(5)−log⁡(4.99)0.01≈0.1990\frac{\log(5) - \log(4.99)}{0.01} \approx 0.19900.01log(5)−log(4.99)​≈0.1990

Flashcard 10: What is the formula for the forward difference quotient to estimate the derivative at a point?

Answer: f′(a)≈f(a+h)−f(a)h.Usesthepointafterf'(a) \approx \frac{f(a+h) - f(a)}{h}. Uses the point after f′(a)≈hf(a+h)−f(a)​.Usesthepointaftera$ to estimate the slope.

Flashcard 11: Estimate f′(0)f'(0)f′(0) for f(x)=cos⁡xf(x) = \cos xf(x)=cosx using h=0.01h = 0.01h=0.01 and the forward difference quotient.

Answer: -0.0005. Forward formula: cos⁡(0.01)−cos⁡(0)0.01≈−0.0005\frac{\cos(0.01) - \cos(0)}{0.01} \approx -0.00050.01cos(0.01)−cos(0)​≈−0.0005

Flashcard 12: What is the impact of a smaller hhh on the accuracy of a derivative estimate?

Answer: Increases accuracy. Smaller step sizes give more precise estimates.

Flashcard 13: Estimate f′(0)f'(0)f′(0) for f(x)=11+xf(x) = \frac{1}{1+x}f(x)=1+x1​ using h=0.1h = 0.1h=0.1 and the symmetric difference quotient.

Answer: -0.9990. Symmetric formula: 1/1.1−1/0.90.2≈−0.9990\frac{1/1.1 - 1/0.9}{0.2} \approx -0.99900.21/1.1−1/0.9​≈−0.9990

Flashcard 14: Estimate f′(1)f'(1)f′(1) for f(x)=x⋅exf(x) = x \cdot e^xf(x)=x⋅ex using h=0.001h = 0.001h=0.001 and the forward difference quotient.

Answer: 5.4379. Forward formula: 1.001⋅e1.001−e0.001≈5.4379\frac{1.001 \cdot e^{1.001} - e}{0.001} \approx 5.43790.0011.001⋅e1.001−e​≈5.4379

Flashcard 15: Estimate f′(2)f'(2)f′(2) for f(x)=ln⁡(1+x2)f(x) = \ln(1+x^2)f(x)=ln(1+x2) using h=0.1h = 0.1h=0.1 and the symmetric difference quotient.

Answer: 0.8000. Symmetric formula: ln⁡(5.41)−ln⁡(4.01)0.2≈0.8000\frac{\ln(5.41) - \ln(4.01)}{0.2} \approx 0.80000.2ln(5.41)−ln(4.01)​≈0.8000

Flashcard 16: Estimate f′(3)f'(3)f′(3) for f(x)=x2+xf(x) = x^2 + xf(x)=x2+x using h=0.1h = 0.1h=0.1 and the symmetric difference quotient.

Answer: 7.0. Symmetric formula gives exact derivative 2x+1=72x+1=72x+1=7 at x=3x=3x=3.

Flashcard 17: Estimate f′(0)f'(0)f′(0) for f(x)=ln⁡(1+x)f(x) = \ln(1+x)f(x)=ln(1+x) using h=0.1h = 0.1h=0.1 and the symmetric difference quotient.

Answer: 0.9950. Symmetric formula: ln⁡(1.1)−ln⁡(0.9)0.2≈0.9950\frac{\ln(1.1) - \ln(0.9)}{0.2} \approx 0.99500.2ln(1.1)−ln(0.9)​≈0.9950

Flashcard 18: Which difference quotient is more accurate: forward or symmetric?

Answer: Symmetric. Averages left and right slopes for better approximation.

Flashcard 19: Estimate f′(1)f'(1)f′(1) for f(x)=exf(x) = e^xf(x)=ex using h=0.001h = 0.001h=0.001 and the backward difference quotient.

Answer: 2.718. Backward formula: e1−e0.9990.001≈2.718\frac{e^1 - e^{0.999}}{0.001} \approx 2.7180.001e1−e0.999​≈2.718

Flashcard 20: Estimate f′(3)f'(3)f′(3) for f(x)=sin⁡xf(x) = \sin xf(x)=sinx using h=0.01h = 0.01h=0.01 and the forward difference quotient.

Answer: -0.9895. Forward formula: sin⁡(3.01)−sin⁡(3)0.01≈−0.9895\frac{\sin(3.01) - \sin(3)}{0.01} \approx -0.98950.01sin(3.01)−sin(3)​≈−0.9895

Flashcard 21: Estimate f′(4)f'(4)f′(4) for f(x)=1x2f(x) = \frac{1}{x^2}f(x)=x21​ using h=0.1h = 0.1h=0.1 and the backward difference quotient.

Answer: -0.1251. Backward formula: 1/16−1/15.210.1≈−0.1251\frac{1/16 - 1/15.21}{0.1} \approx -0.12510.11/16−1/15.21​≈−0.1251

Flashcard 22: Estimate f′(3)f'(3)f′(3) for f(x)=cos⁡(2x)f(x) = \cos(2x)f(x)=cos(2x) using h=0.01h = 0.01h=0.01 and the backward difference quotient.

Answer: -1.9159. Backward formula: cos⁡(6)−cos⁡(5.98)0.01≈−1.9159\frac{\cos(6) - \cos(5.98)}{0.01} \approx -1.91590.01cos(6)−cos(5.98)​≈−1.9159

Flashcard 23: Estimate f′(2)f'(2)f′(2) for f(x)=1+xf(x) = \sqrt{1+x}f(x)=1+x​ using h=0.1h = 0.1h=0.1 and the symmetric difference quotient.

Answer: 0.3536. Symmetric formula: 3.1−2.90.2≈0.3536\frac{\sqrt{3.1} - \sqrt{2.9}}{0.2} \approx 0.35360.23.1​−2.9​​≈0.3536

Flashcard 24: Estimate f′(2)f'(2)f′(2) for f(x)=x3−3x2+2xf(x) = x^3 - 3x^2 + 2xf(x)=x3−3x2+2x using h=0.1h = 0.1h=0.1 and the forward difference quotient.

Answer: -2.999. Forward formula for f′(x)=3x2−6x+2f'(x) = 3x^2 - 6x + 2f′(x)=3x2−6x+2 at x=2x=2x=2.

Flashcard 25: Which value of hhh gives a more accurate estimate: h=0.1h=0.1h=0.1 or h=0.01h=0.01h=0.01?

Answer: h=0.01h=0.01h=0.01. Smaller hhh values reduce approximation error.

Flashcard 26: Estimate f′(4)f'(4)f′(4) for f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ using h=0.1h = 0.1h=0.1 and the forward difference quotient.

Answer: -0.0625. Forward formula: 1/4.1−1/40.1≈−0.0625\frac{1/4.1 - 1/4}{0.1} \approx -0.06250.11/4.1−1/4​≈−0.0625

Flashcard 27: What is the formula to estimate the derivative of a function at a point using the symmetric difference quotient?

Answer: f′(a)≈f(a+h)−f(a−h)2hf'(a) \approx \frac{f(a+h) - f(a-h)}{2h}f′(a)≈2hf(a+h)−f(a−h)​. Uses points on both sides of aaa for better accuracy.

Flashcard 28: Identify the error: f′(a)≈f(a+h)+f(a−h)2hf'(a) \approx \frac{f(a+h) + f(a-h)}{2h}f′(a)≈2hf(a+h)+f(a−h)​

Answer: Correct: f′(a)≈f(a+h)−f(a−h)2hf'(a) \approx \frac{f(a+h) - f(a-h)}{2h}f′(a)≈2hf(a+h)−f(a−h)​. Should subtract, not add, in the numerator.

Flashcard 29: Estimate f′(1)f'(1)f′(1) for f(x)=e−xf(x) = e^{-x}f(x)=e−x using h=0.001h = 0.001h=0.001 and the backward difference quotient.

Answer: -0.368. Backward formula: e−1−e−1.0010.001≈−0.368\frac{e^{-1} - e^{-1.001}}{0.001} \approx -0.3680.001e−1−e−1.001​≈−0.368

Flashcard 30: Estimate f′(2)f'(2)f′(2) for f(x)=x2f(x) = x^2f(x)=x2 using h=0.01h = 0.01h=0.01 and the symmetric difference quotient.

Answer: 4.0001. Symmetric formula: (2.01)2−(1.99)20.02=4.0001\frac{(2.01)^2 - (1.99)^2}{0.02} = 4.00010.02(2.01)2−(1.99)2​=4.0001