AP Calculus BC Flashcards: Estimating Derivatives Of A Function

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.

Flashcard 1: Estimate $f'(2)$ for $f(x) = x^3 - 3x^2 + 2x$ using $h = 0.1$ and the forward difference quotient.

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

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

Answer: -1.9159. Backward formula: $\frac{\cos(6) - \cos(5.98)}{0.01} \approx -1.9159$

Flashcard 3: Estimate $f'(3)$ for $f(x) = \sin x$ using $h = 0.01$ and the forward difference quotient.

Answer: -0.9895. Forward formula: $\frac{\sin(3.01) - \sin(3)}{0.01} \approx -0.9895$

Flashcard 4: Identify the error: $f'(a) \approx \frac{f(a+h) + f(a-h)}{2h}$

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

Flashcard 5: Which difference quotient is more accurate: forward or symmetric?

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

Flashcard 6: Estimate $f'(1)$ for $f(x) = e^x$ using $h = 0.001$ and the backward difference quotient.

Answer: 2.718. Backward formula: $\frac{e^1 - e^{0.999}}{0.001} \approx 2.718$

Flashcard 7: Estimate $f'(1)$ for $f(x) = e^{-x}$ using $h = 0.001$ and the backward difference quotient.

Answer: -0.368. Backward formula: $\frac{e^{-1} - e^{-1.001}}{0.001} \approx -0.368$

Flashcard 8: For $f(x) = x^3$, estimate $f'(2)$ using $h = 0.1$ and the symmetric difference quotient.

Answer: 12.01. Symmetric formula: $\frac{(2.1)^3 - (1.9)^3}{0.2} = 12.01$

Flashcard 9: Estimate $f'(0)$ for $f(x) = \sin(x^2)$ using $h = 0.01$ and the backward difference quotient.

Answer: 0.0000. Backward formula: derivative is 0 at $x=0$ for even functions.

Flashcard 10: Estimate $f'(3)$ for $f(x) = \cos(2x)$ using $h = 0.01$ and the backward difference quotient.

Answer: -1.9159. Backward formula: $\frac{\cos(6) - \cos(5.98)}{0.01} \approx -1.9159$

Flashcard 11: What is the formula to estimate the derivative of a function at a point using the symmetric difference quotient?

Answer: $f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}$. Uses points on both sides of $a$ for better accuracy.

Flashcard 12: Estimate $f'(3)$ for $f(x) = x^2 + x$ using $h = 0.1$ and the symmetric difference quotient.

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

Flashcard 13: Estimate $f'(5)$ for $f(x) = \log x$ using $h = 0.01$ and the backward difference quotient.

Answer: 0.1990. Backward formula: $\frac{\log(5) - \log(4.99)}{0.01} \approx 0.1990$

Flashcard 14: Estimate $f'(4)$ for $f(x) = \frac{1}{x^3}$ using $h = 0.1$ and the forward difference quotient.

Answer: -0.0351. Forward formula: $\frac{1/(4.1)^3 - 1/64}{0.1} \approx -0.0351$

Flashcard 15: Estimate $f'(1)$ for $f(x) = \sin(2x)$ using $h = 0.001$ and the symmetric difference quotient.

Answer: 1.0806. Symmetric formula: $\frac{\sin(2.002) - \sin(1.998)}{0.002} \approx 1.0806$

Flashcard 16: Estimate $f'(4)$ for $f(x) = \frac{1}{x^2}$ using $h = 0.1$ and the backward difference quotient.

Answer: -0.1251. Backward formula: $\frac{1/16 - 1/15.21}{0.1} \approx -0.1251$

Flashcard 17: Estimate $f'(3)$ for $f(x) = x^2 + 3x$ using $h = 0.1$ and the forward difference quotient.

Answer: 9.0. Forward formula gives exact derivative $2x+3=9$ at $x=3$.

Flashcard 18: Estimate $f'(0)$ for $f(x) = \cos x$ using $h = 0.01$ and the forward difference quotient.

Answer: -0.0005. Forward formula: $\frac{\cos(0.01) - \cos(0)}{0.01} \approx -0.0005$

Flashcard 19: What is the formula for the forward difference quotient to estimate the derivative at a point?

Answer: $f'(a) \approx \frac{f(a+h) - f(a)}{h}$. Uses the point after $a$ to estimate the slope.

Flashcard 20: Estimate $f'(1)$ for $f(x) = \sqrt{x}$ using $h = 0.001$ and the backward difference quotient.

Answer: 0.4995. Backward formula: $\frac{\sqrt{1} - \sqrt{0.999}}{0.001} \approx 0.4995$

Flashcard 21: Estimate $f'(2)$ for $f(x) = \ln(1+x^2)$ using $h = 0.1$ and the symmetric difference quotient.

Answer: 0.8000. Symmetric formula: $\frac{\ln(5.41) - \ln(4.01)}{0.2} \approx 0.8000$

Flashcard 22: For $f(x) = x^4$, estimate $f'(1)$ using $h = 0.1$ and the forward difference quotient.

Answer: 4.0301. Forward formula: $\frac{(1.1)^4 - 1}{0.1} \approx 4.0301$

Flashcard 23: Which difference quotient is more accurate: forward or symmetric?

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

Flashcard 24: What is the impact of a smaller $h$ on the accuracy of a derivative estimate?

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

Flashcard 25: Estimate $f'(2)$ for $f(x) = \tan x$ using $h = 0.01$ and the forward difference quotient.

Answer: 2.0199. Forward formula: $\frac{\tan(2.01) - \tan(2)}{0.01} \approx 2.0199$

Flashcard 26: Estimate $f'(0)$ for $f(x) = e^{2x}$ using $h = 0.01$ and the symmetric difference quotient.

Answer: 2.0000. Symmetric formula: $\frac{e^{0.02} - e^{-0.02}}{0.02} \approx 2.0000$

Flashcard 27: Identify the error: $f'(a) \approx \frac{f(a+h) + f(a-h)}{2h}$

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

Flashcard 28: Estimate $f'(0)$ for $f(x) = \cos x$ using $h = 0.01$ and the forward difference quotient.

Answer: -0.0005. Forward formula: $\frac{\cos(0.01) - \cos(0)}{0.01} \approx -0.0005$

Flashcard 29: Estimate $f'(2)$ for $f(x) = x^3 - 3x^2 + 2x$ using $h = 0.1$ and the forward difference quotient.

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

Flashcard 30: Which value of $h$ gives a more accurate estimate: $h=0.1$ or $h=0.01$?

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

Flashcard 31: Estimate $f'(3)$ for $f(x) = \tan^{-1}(x)$ using $h = 0.01$ and the backward difference quotient.

Answer: 0.0995. Backward formula: $\frac{\tan^{-1}(3) - \tan^{-1}(2.99)}{0.01} \approx 0.0995$

Flashcard 32: Estimate $f'(0)$ for $f(x) = \frac{1}{1+x}$ using $h = 0.1$ and the symmetric difference quotient.

Answer: -0.9990. Symmetric formula: $\frac{1/1.1 - 1/0.9}{0.2} \approx -0.9990$

Flashcard 33: Estimate $f'(2)$ for $f(x) = \sqrt{1+x}$ using $h = 0.1$ and the symmetric difference quotient.

Answer: 0.3536. Symmetric formula: $\frac{\sqrt{3.1} - \sqrt{2.9}}{0.2} \approx 0.3536$

Flashcard 34: Estimate $f'(2)$ for $f(x) = \ln x$ using $h = 0.01$ and the backward difference quotient.

Answer: 0.5001. Backward formula: $\frac{\ln(2) - \ln(1.99)}{0.01} \approx 0.5001$

Flashcard 35: Estimate $f'(0)$ for $f(x) = \log(1+x)$ using $h = 0.01$ and the symmetric difference quotient.

Answer: 0.9950. Symmetric formula: $\frac{\log(1.01) - \log(0.99)}{0.02} \approx 0.9950$

Flashcard 36: Which value of $h$ gives a more accurate estimate: $h=0.1$ or $h=0.01$?

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

Flashcard 37: Estimate $f'(0)$ for $f(x) = \ln(1+x)$ using $h = 0.1$ and the symmetric difference quotient.

Answer: 0.9950. Symmetric formula: $\frac{\ln(1.1) - \ln(0.9)}{0.2} \approx 0.9950$

Flashcard 38: Estimate $f'(3)$ for $f(x) = x^2 + x$ using $h = 0.1$ and the symmetric difference quotient.

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

Flashcard 39: Estimate $f'(3)$ for $f(x) = \sin x$ using $h = 0.01$ and the forward difference quotient.

Answer: -0.9895. Forward formula: $\frac{\sin(3.01) - \sin(3)}{0.01} \approx -0.9895$

Flashcard 40: Estimate $f'(5)$ for $f(x) = x^2$ using $h = 0.01$ and the forward difference quotient.

Answer: 10.01. Forward formula: $\frac{(5.01)^2 - 25}{0.01} = 10.01$

Flashcard 41: Estimate $f'(2)$ for $f(x) = \sqrt{1+x}$ using $h = 0.1$ and the symmetric difference quotient.

Answer: 0.3536. Symmetric formula: $\frac{\sqrt{3.1} - \sqrt{2.9}}{0.2} \approx 0.3536$

Flashcard 42: Estimate $f'(0)$ for $f(x) = \log(1+x)$ using $h = 0.01$ and the symmetric difference quotient.

Answer: 0.9950. Symmetric formula: $\frac{\log(1.01) - \log(0.99)}{0.02} \approx 0.9950$

Flashcard 43: Estimate $f'(1)$ for $f(x) = x \cdot e^x$ using $h = 0.001$ and the forward difference quotient.

Answer: 5.4379. Forward formula: $\frac{1.001 \cdot e^{1.001} - e}{0.001} \approx 5.4379$

Flashcard 44: Estimate $f'(2)$ for $f(x) = x^2$ using $h = 0.01$ and the symmetric difference quotient.

Answer: 4.0001. Symmetric formula: $\frac{(2.01)^2 - (1.99)^2}{0.02} = 4.0001$

Flashcard 45: Estimate $f'(1)$ for $f(x) = e^{-x}$ using $h = 0.001$ and the backward difference quotient.

Answer: -0.368. Backward formula: $\frac{e^{-1} - e^{-1.001}}{0.001} \approx -0.368$

Flashcard 46: What is the formula for the backward difference quotient to estimate the derivative at a point?

Answer: $f'(a) \approx \frac{f(a) - f(a-h)}{h}$. Uses the point before $a$ to estimate the slope.

Flashcard 47: Estimate $f'(1)$ for $f(x) = 2^x$ using $h = 0.001$ and the symmetric difference quotient.

Answer: 1.3863. Symmetric formula: $\frac{2^{1.001} - 2^{0.999}}{0.002} \approx 1.3863$

Flashcard 48: Estimate $f'(0)$ for $f(x) = \ln(1+x)$ using $h = 0.1$ and the symmetric difference quotient.

Answer: 0.9950. Symmetric formula: $\frac{\ln(1.1) - \ln(0.9)}{0.2} \approx 0.9950$

Flashcard 49: What is the formula for the backward difference quotient to estimate the derivative at a point?

Answer: $f'(a) \approx \frac{f(a) - f(a-h)}{h}$. Uses the point before $a$ to estimate the slope.

Flashcard 50: For $f(x) = x^3$, estimate $f'(2)$ using $h = 0.1$ and the symmetric difference quotient.

Answer: 12.01. Symmetric formula: $\frac{(2.1)^3 - (1.9)^3}{0.2} = 12.01$

Flashcard 51: What is the impact of a smaller $h$ on the accuracy of a derivative estimate?

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

Flashcard 52: Estimate $f'(0)$ for $f(x) = \frac{1}{1+x}$ using $h = 0.1$ and the symmetric difference quotient.

Answer: -0.9990. Symmetric formula: $\frac{1/1.1 - 1/0.9}{0.2} \approx -0.9990$

Flashcard 53: Estimate $f'(1)$ for $f(x) = \sqrt{x}$ using $h = 0.001$ and the backward difference quotient.

Answer: 0.4995. Backward formula: $\frac{\sqrt{1} - \sqrt{0.999}}{0.001} \approx 0.4995$

Flashcard 54: Which method reduces error due to truncation: forward or symmetric difference quotient?

Answer: Symmetric. Balances left and right approximation errors.

Flashcard 55: Estimate $f'(5)$ for $f(x) = \log x$ using $h = 0.01$ and the backward difference quotient.

Answer: 0.1990. Backward formula: $\frac{\log(5) - \log(4.99)}{0.01} \approx 0.1990$

Flashcard 56: Estimate $f'(2)$ for $f(x) = \tan x$ using $h = 0.01$ and the forward difference quotient.

Answer: 2.0199. Forward formula: $\frac{\tan(2.01) - \tan(2)}{0.01} \approx 2.0199$

Flashcard 57: Estimate $f'(0)$ for $f(x) = \sin(x^2)$ using $h = 0.01$ and the backward difference quotient.

Answer: 0.0000. Backward formula: derivative is 0 at $x=0$ for even functions.

Flashcard 58: Estimate $f'(1)$ for $f(x) = 2^x$ using $h = 0.001$ and the symmetric difference quotient.

Answer: 1.3863. Symmetric formula: $\frac{2^{1.001} - 2^{0.999}}{0.002} \approx 1.3863$

Flashcard 59: Estimate $f'(5)$ for $f(x) = x^2$ using $h = 0.01$ and the forward difference quotient.

Answer: 10.01. Forward formula: $\frac{(5.01)^2 - 25}{0.01} = 10.01$

Flashcard 60: Which method reduces error due to truncation: forward or symmetric difference quotient?

Answer: Symmetric. Balances left and right approximation errors.

Flashcard 61: Estimate $f'(1)$ for $f(x) = \sin(2x)$ using $h = 0.001$ and the symmetric difference quotient.

Answer: 1.0806. Symmetric formula: $\frac{\sin(2.002) - \sin(1.998)}{0.002} \approx 1.0806$

Flashcard 62: Estimate $f'(4)$ for $f(x) = \frac{1}{x}$ using $h = 0.1$ and the forward difference quotient.

Answer: -0.0625. Forward formula: $\frac{1/4.1 - 1/4}{0.1} \approx -0.0625$

Flashcard 63: Estimate $f'(0)$ for $f(x) = e^{2x}$ using $h = 0.01$ and the symmetric difference quotient.

Answer: 2.0000. Symmetric formula: $\frac{e^{0.02} - e^{-0.02}}{0.02} \approx 2.0000$

Flashcard 64: Estimate $f'(3)$ for $f(x) = \tan^{-1}(x)$ using $h = 0.01$ and the backward difference quotient.

Answer: 0.0995. Backward formula: $\frac{\tan^{-1}(3) - \tan^{-1}(2.99)}{0.01} \approx 0.0995$

Flashcard 65: What is the formula for the forward difference quotient to estimate the derivative at a point?

Answer: $f'(a) \approx \frac{f(a+h) - f(a)}{h}. Uses the point after$a$ to estimate the slope.

Flashcard 66: Estimate $f'(1)$ for $f(x) = x \cdot e^x$ using $h = 0.001$ and the forward difference quotient.

Answer: 5.4379. Forward formula: $\frac{1.001 \cdot e^{1.001} - e}{0.001} \approx 5.4379$

Flashcard 67: Estimate $f'(3)$ for $f(x) = x^2 + 3x$ using $h = 0.1$ and the forward difference quotient.

Answer: 9.0. Forward formula gives exact derivative $2x+3=9$ at $x=3$.

Flashcard 68: Estimate $f'(2)$ for $f(x) = x^2$ using $h = 0.01$ and the symmetric difference quotient.

Answer: 4.0001. Symmetric formula: $\frac{(2.01)^2 - (1.99)^2}{0.02} = 4.0001$

Flashcard 69: Estimate $f'(4)$ for $f(x) = \frac{1}{x^3}$ using $h = 0.1$ and the forward difference quotient.

Answer: -0.0351. Forward formula: $\frac{1/(4.1)^3 - 1/64}{0.1} \approx -0.0351$

Flashcard 70: For $f(x) = x^4$, estimate $f'(1)$ using $h = 0.1$ and the forward difference quotient.

Answer: 4.0301. Forward formula: $\frac{(1.1)^4 - 1}{0.1} \approx 4.0301$

Flashcard 71: Estimate $f'(4)$ for $f(x) = \frac{1}{x}$ using $h = 0.1$ and the forward difference quotient.

Answer: -0.0625. Forward formula: $\frac{1/4.1 - 1/4}{0.1} \approx -0.0625$

Flashcard 72: Estimate $f'(2)$ for $f(x) = \ln(1+x^2)$ using $h = 0.1$ and the symmetric difference quotient.

Answer: 0.8000. Symmetric formula: $\frac{\ln(5.41) - \ln(4.01)}{0.2} \approx 0.8000$

Flashcard 73: Estimate $f'(1)$ for $f(x) = e^x$ using $h = 0.001$ and the backward difference quotient.

Answer: 2.718. Backward formula: $\frac{e^1 - e^{0.999}}{0.001} \approx 2.718$

Flashcard 74: What is the formula to estimate the derivative of a function at a point using the symmetric difference quotient?

Answer: $f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}$. Uses points on both sides of $a$ for better accuracy.