Estimating Derivatives of a Function - AP Calculus BC
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What is the formula for the backward difference quotient to estimate the derivative at a point?
What is the formula for the backward difference quotient to estimate the derivative at a point?
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$f'(a) \approx \frac{f(a) - f(a-h)}{h}$. Uses the point before $a$ to estimate the slope.
$f'(a) \approx \frac{f(a) - f(a-h)}{h}$. Uses the point before $a$ to estimate the slope.
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Estimate $f'(3)$ for $f(x) = \tan^{-1}(x)$ using $h = 0.01$ and the backward difference quotient.
Estimate $f'(3)$ for $f(x) = \tan^{-1}(x)$ using $h = 0.01$ and the backward difference quotient.
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0.0995. Backward formula: $\frac{\tan^{-1}(3) - \tan^{-1}(2.99)}{0.01} \approx 0.0995$
0.0995. Backward formula: $\frac{\tan^{-1}(3) - \tan^{-1}(2.99)}{0.01} \approx 0.0995$
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Estimate $f'(1)$ for $f(x) = 2^x$ using $h = 0.001$ and the symmetric difference quotient.
Estimate $f'(1)$ for $f(x) = 2^x$ using $h = 0.001$ and the symmetric difference quotient.
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1.3863. Symmetric formula: $\frac{2^{1.001} - 2^{0.999}}{0.002} \approx 1.3863$
1.3863. Symmetric formula: $\frac{2^{1.001} - 2^{0.999}}{0.002} \approx 1.3863$
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Estimate $f'(0)$ for $f(x) = \log(1+x)$ using $h = 0.01$ and the symmetric difference quotient.
Estimate $f'(0)$ for $f(x) = \log(1+x)$ using $h = 0.01$ and the symmetric difference quotient.
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0.9950. Symmetric formula: $\frac{\log(1.01) - \log(0.99)}{0.02} \approx 0.9950$
0.9950. Symmetric formula: $\frac{\log(1.01) - \log(0.99)}{0.02} \approx 0.9950$
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Estimate $f'(4)$ for $f(x) = \frac{1}{x^3}$ using $h = 0.1$ and the forward difference quotient.
Estimate $f'(4)$ for $f(x) = \frac{1}{x^3}$ using $h = 0.1$ and the forward difference quotient.
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-0.0351. Forward formula: $\frac{1/(4.1)^3 - 1/64}{0.1} \approx -0.0351$
-0.0351. Forward formula: $\frac{1/(4.1)^3 - 1/64}{0.1} \approx -0.0351$
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Estimate $f'(1)$ for $f(x) = \sin(2x)$ using $h = 0.001$ and the symmetric difference quotient.
Estimate $f'(1)$ for $f(x) = \sin(2x)$ using $h = 0.001$ and the symmetric difference quotient.
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1.0806. Symmetric formula: $\frac{\sin(2.002) - \sin(1.998)}{0.002} \approx 1.0806$
1.0806. Symmetric formula: $\frac{\sin(2.002) - \sin(1.998)}{0.002} \approx 1.0806$
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Estimate $f'(2)$ for $f(x) = \tan x$ using $h = 0.01$ and the forward difference quotient.
Estimate $f'(2)$ for $f(x) = \tan x$ using $h = 0.01$ and the forward difference quotient.
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2.0199. Forward formula: $\frac{\tan(2.01) - \tan(2)}{0.01} \approx 2.0199$
2.0199. Forward formula: $\frac{\tan(2.01) - \tan(2)}{0.01} \approx 2.0199$
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Estimate $f'(1)$ for $f(x) = \sqrt{x}$ using $h = 0.001$ and the backward difference quotient.
Estimate $f'(1)$ for $f(x) = \sqrt{x}$ using $h = 0.001$ and the backward difference quotient.
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0.4995. Backward formula: $\frac{\sqrt{1} - \sqrt{0.999}}{0.001} \approx 0.4995$
0.4995. Backward formula: $\frac{\sqrt{1} - \sqrt{0.999}}{0.001} \approx 0.4995$
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Estimate $f'(5)$ for $f(x) = \log x$ using $h = 0.01$ and the backward difference quotient.
Estimate $f'(5)$ for $f(x) = \log x$ using $h = 0.01$ and the backward difference quotient.
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0.1990. Backward formula: $\frac{\log(5) - \log(4.99)}{0.01} \approx 0.1990$
0.1990. Backward formula: $\frac{\log(5) - \log(4.99)}{0.01} \approx 0.1990$
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What is the formula for the forward difference quotient to estimate the derivative at a point?
What is the formula for the forward difference quotient to estimate the derivative at a point?
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$f'(a) \approx \frac{f(a+h) - f(a)}{h}. Uses the point after $a$ to estimate the slope.
$f'(a) \approx \frac{f(a+h) - f(a)}{h}. Uses the point after $a$ to estimate the slope.
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Estimate $f'(0)$ for $f(x) = \cos x$ using $h = 0.01$ and the forward difference quotient.
Estimate $f'(0)$ for $f(x) = \cos x$ using $h = 0.01$ and the forward difference quotient.
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-0.0005. Forward formula: $\frac{\cos(0.01) - \cos(0)}{0.01} \approx -0.0005$
-0.0005. Forward formula: $\frac{\cos(0.01) - \cos(0)}{0.01} \approx -0.0005$
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What is the impact of a smaller $h$ on the accuracy of a derivative estimate?
What is the impact of a smaller $h$ on the accuracy of a derivative estimate?
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Increases accuracy. Smaller step sizes give more precise estimates.
Increases accuracy. Smaller step sizes give more precise estimates.
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Estimate $f'(0)$ for $f(x) = \frac{1}{1+x}$ using $h = 0.1$ and the symmetric difference quotient.
Estimate $f'(0)$ for $f(x) = \frac{1}{1+x}$ using $h = 0.1$ and the symmetric difference quotient.
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-0.9990. Symmetric formula: $\frac{1/1.1 - 1/0.9}{0.2} \approx -0.9990$
-0.9990. Symmetric formula: $\frac{1/1.1 - 1/0.9}{0.2} \approx -0.9990$
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Estimate $f'(1)$ for $f(x) = x \cdot e^x$ using $h = 0.001$ and the forward difference quotient.
Estimate $f'(1)$ for $f(x) = x \cdot e^x$ using $h = 0.001$ and the forward difference quotient.
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5.4379. Forward formula: $\frac{1.001 \cdot e^{1.001} - e}{0.001} \approx 5.4379$
5.4379. Forward formula: $\frac{1.001 \cdot e^{1.001} - e}{0.001} \approx 5.4379$
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Estimate $f'(2)$ for $f(x) = \ln(1+x^2)$ using $h = 0.1$ and the symmetric difference quotient.
Estimate $f'(2)$ for $f(x) = \ln(1+x^2)$ using $h = 0.1$ and the symmetric difference quotient.
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0.8000. Symmetric formula: $\frac{\ln(5.41) - \ln(4.01)}{0.2} \approx 0.8000$
0.8000. Symmetric formula: $\frac{\ln(5.41) - \ln(4.01)}{0.2} \approx 0.8000$
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Estimate $f'(3)$ for $f(x) = x^2 + x$ using $h = 0.1$ and the symmetric difference quotient.
Estimate $f'(3)$ for $f(x) = x^2 + x$ using $h = 0.1$ and the symmetric difference quotient.
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7.0. Symmetric formula gives exact derivative $2x+1=7$ at $x=3$.
7.0. Symmetric formula gives exact derivative $2x+1=7$ at $x=3$.
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Estimate $f'(0)$ for $f(x) = \ln(1+x)$ using $h = 0.1$ and the symmetric difference quotient.
Estimate $f'(0)$ for $f(x) = \ln(1+x)$ using $h = 0.1$ and the symmetric difference quotient.
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0.9950. Symmetric formula: $\frac{\ln(1.1) - \ln(0.9)}{0.2} \approx 0.9950$
0.9950. Symmetric formula: $\frac{\ln(1.1) - \ln(0.9)}{0.2} \approx 0.9950$
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Which difference quotient is more accurate: forward or symmetric?
Which difference quotient is more accurate: forward or symmetric?
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Symmetric. Averages left and right slopes for better approximation.
Symmetric. Averages left and right slopes for better approximation.
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Estimate $f'(1)$ for $f(x) = e^x$ using $h = 0.001$ and the backward difference quotient.
Estimate $f'(1)$ for $f(x) = e^x$ using $h = 0.001$ and the backward difference quotient.
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2.718. Backward formula: $\frac{e^1 - e^{0.999}}{0.001} \approx 2.718$
2.718. Backward formula: $\frac{e^1 - e^{0.999}}{0.001} \approx 2.718$
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Estimate $f'(3)$ for $f(x) = \sin x$ using $h = 0.01$ and the forward difference quotient.
Estimate $f'(3)$ for $f(x) = \sin x$ using $h = 0.01$ and the forward difference quotient.
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-0.9895. Forward formula: $\frac{\sin(3.01) - \sin(3)}{0.01} \approx -0.9895$
-0.9895. Forward formula: $\frac{\sin(3.01) - \sin(3)}{0.01} \approx -0.9895$
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Estimate $f'(4)$ for $f(x) = \frac{1}{x^2}$ using $h = 0.1$ and the backward difference quotient.
Estimate $f'(4)$ for $f(x) = \frac{1}{x^2}$ using $h = 0.1$ and the backward difference quotient.
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-0.1251. Backward formula: $\frac{1/16 - 1/15.21}{0.1} \approx -0.1251$
-0.1251. Backward formula: $\frac{1/16 - 1/15.21}{0.1} \approx -0.1251$
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Estimate $f'(3)$ for $f(x) = \cos(2x)$ using $h = 0.01$ and the backward difference quotient.
Estimate $f'(3)$ for $f(x) = \cos(2x)$ using $h = 0.01$ and the backward difference quotient.
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-1.9159. Backward formula: $\frac{\cos(6) - \cos(5.98)}{0.01} \approx -1.9159$
-1.9159. Backward formula: $\frac{\cos(6) - \cos(5.98)}{0.01} \approx -1.9159$
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Estimate $f'(2)$ for $f(x) = \sqrt{1+x}$ using $h = 0.1$ and the symmetric difference quotient.
Estimate $f'(2)$ for $f(x) = \sqrt{1+x}$ using $h = 0.1$ and the symmetric difference quotient.
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0.3536. Symmetric formula: $\frac{\sqrt{3.1} - \sqrt{2.9}}{0.2} \approx 0.3536$
0.3536. Symmetric formula: $\frac{\sqrt{3.1} - \sqrt{2.9}}{0.2} \approx 0.3536$
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Estimate $f'(2)$ for $f(x) = x^3 - 3x^2 + 2x$ using $h = 0.1$ and the forward difference quotient.
Estimate $f'(2)$ for $f(x) = x^3 - 3x^2 + 2x$ using $h = 0.1$ and the forward difference quotient.
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-2.999. Forward formula for $f'(x) = 3x^2 - 6x + 2$ at $x=2$.
-2.999. Forward formula for $f'(x) = 3x^2 - 6x + 2$ at $x=2$.
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Which value of $h$ gives a more accurate estimate: $h=0.1$ or $h=0.01$?
Which value of $h$ gives a more accurate estimate: $h=0.1$ or $h=0.01$?
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$h=0.01$. Smaller $h$ values reduce approximation error.
$h=0.01$. Smaller $h$ values reduce approximation error.
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Estimate $f'(4)$ for $f(x) = \frac{1}{x}$ using $h = 0.1$ and the forward difference quotient.
Estimate $f'(4)$ for $f(x) = \frac{1}{x}$ using $h = 0.1$ and the forward difference quotient.
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-0.0625. Forward formula: $\frac{1/4.1 - 1/4}{0.1} \approx -0.0625$
-0.0625. Forward formula: $\frac{1/4.1 - 1/4}{0.1} \approx -0.0625$
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What is the formula to estimate the derivative of a function at a point using the symmetric difference quotient?
What is the formula to estimate the derivative of a function at a point using the symmetric difference quotient?
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$f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}$. Uses points on both sides of $a$ for better accuracy.
$f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}$. Uses points on both sides of $a$ for better accuracy.
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Identify the error: $f'(a) \approx \frac{f(a+h) + f(a-h)}{2h}$
Identify the error: $f'(a) \approx \frac{f(a+h) + f(a-h)}{2h}$
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Correct: $f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}$. Should subtract, not add, in the numerator.
Correct: $f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}$. Should subtract, not add, in the numerator.
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Estimate $f'(1)$ for $f(x) = e^{-x}$ using $h = 0.001$ and the backward difference quotient.
Estimate $f'(1)$ for $f(x) = e^{-x}$ using $h = 0.001$ and the backward difference quotient.
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-0.368. Backward formula: $\frac{e^{-1} - e^{-1.001}}{0.001} \approx -0.368$
-0.368. Backward formula: $\frac{e^{-1} - e^{-1.001}}{0.001} \approx -0.368$
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Estimate $f'(2)$ for $f(x) = x^2$ using $h = 0.01$ and the symmetric difference quotient.
Estimate $f'(2)$ for $f(x) = x^2$ using $h = 0.01$ and the symmetric difference quotient.
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4.0001. Symmetric formula: $\frac{(2.01)^2 - (1.99)^2}{0.02} = 4.0001$
4.0001. Symmetric formula: $\frac{(2.01)^2 - (1.99)^2}{0.02} = 4.0001$
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