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Calculus 1 : How to find differential functions

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Example Questions

Example Question #551 : How To Find Differential Functions

Find the derivative of $g(x) = \sqrt{x^3+5}$

Possible Answers:

$\frac{6x}{2\sqrt{x^3+5}}$

$\frac{3x^2}{2\sqrt{x^3+5}}$

$-3x^2 \sqrt{x^3+5}$

$\frac{1}{2\sqrt{x^3+5}}$

Correct answer:

$\frac{3x^2}{2\sqrt{x^3+5}}$

Explanation:

Rewrite the square root as the $\small \frac{1}{2}$ power to apply the power rule more easily. Then be sure to use the chain rule to fully differentiate.

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Example Question #739 : Differential Functions

Find the derviative of $f(x) = \sqrt{\ln{x}}$

Possible Answers:

$\sqrt{\frac{1}{x}}$

$\small \frac{1}{\sqrt{\ln{x}}}$

$\small \frac{1}{2x \sqrt{\ln{x}}}$

$\frac{1}{2x}$

$- \frac{1}{2x}$

Correct answer:

$\small \frac{1}{2x \sqrt{\ln{x}}}$

Explanation:

Apply the Chain Rule to the function $\small \small f(x) = (\ln{x})^{\frac{1}{2}}$

This gives $\small \frac{1}{2} (\ln{x})^{-\frac{1}{2}} \cdot \frac{1}{x}$ . Simplify.

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Example Question #740 : Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that f'(x)=csc(cot(x)) and y(2)=4 , approximate y(2.04) using Euler's Method and four steps.

Possible Answers:

4.019

3.996

3.888

3.912

3.749

Correct answer:

3.912

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

To calculate the step size find the distance between the final and initial value and divide by the number of steps to be used:

$\Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told f'(x)=csc(cot(x)) and y(2)=4

Knowing this, we may take the steps to estimate our function value at our final value:

$\Delta x = \frac{2.04-2}{4}=0.01$

y_0=4;x_0=2

y_1=4+(0.01)csc(cot(2))=3.977

y_2=3.977+(0.01)csc(cot(2.01))=3.955

y_3=3.955+(0.01)csc(cot(2.02))=3.933

y_4=3.933+(0.01)csc(cot(2.03))=3.912

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Example Question #741 : Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=sin(2^{e^{x}})$ and y(3)=2 , approximate y(6) using Euler's Method and three steps.

Possible Answers:

0.915

2.147

1.063

1.561

0.383

Correct answer:

1.063

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

To calculate the step size find the distance between the final and initial value and divide by the number of steps to be used:

$\Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $f'(x)=sin(2^{e^{x}})$ and y(3)=2

Knowing this, we may take the steps to estimate our function value at our final value:

$\Delta x = \frac{6-3}{3}=1$

y_0=2,x_0=3

$y_1=2+(1)sin(2^{e^{3}})=1.326$

$y_2=1.326+(1)sin(2^{e^{4}})=0.916$

$y_3=0.916+(1)sin(2^{e^{5}})=1.063$

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Example Question #742 : Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=(cos(x))^{cos(x)}$ and y(0)=4 , approximate y(1.5) using Euler's Method and three steps.

Possible Answers:

5.112

5.305

4.783

5.496

4.292

Correct answer:

5.305

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

To calculate the step size find the distance between the final and initial value and divide by the number of steps to be used:

$\Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $f'(x)=(cos(x))^{cos(x)}$ and y(0)=4

Knowing this, we may take the steps to estimate our function value at our final value:

$\Delta x = \frac{1.5-0}{3}=0.5$

y_0=4,x_0=0

$y_1=4+0.5(cos(0))^{cos(0)}=4.5$

$y_2=4.5+0.5(cos(0.5))^{cos(0.5)}=4.946$

$y_3=4.946+0.5(cos(1))^{cos(1)}=5.305$

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Example Question #743 : Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that f'(x)=cot(x^2) and y(1)=5 , approximate y(1.6) using Euler's Method and three steps.

Possible Answers:

4.938

5.276

5.072

5.191

5.409

Correct answer:

5.072

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

To calculate the step size find the distance between the final and initial value and divide by the number of steps to be used:

$\Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told f'(x)=cot(x^2) and y(1)=5

Knowing this, we may take the steps to estimate our function value at our final value:

$\Delta x = \frac{1.6-1}{3}=0.2$

y_0=5,x_0=1

y_1=5+(0.2)cot(1^2)=5.128

y_2=5.128+(0.2)cot(1.2^2)=5.154

y_3=5.154+(0.2)cot(1.4^2)=5.072

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Example Question #744 : Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that f'(x)=tan(ln(ln(x))) and y(2)=0 , approximate y(5) using Euler's Method and three steps.

Possible Answers:

0.183

0.049

-0.517

-0.384

-0.265

Correct answer:

0.049

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

To calculate the step size find the distance between the final and initial value and divide by the number of steps to be used:

$\Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told f'(x)=tan(ln(ln(x))) and y(2)=0

Knowing this, we may take the steps to estimate our function value at our final value:

$\Delta x = \frac{5-2}{3}=1$

y_0=0;x_0=2

y_1=0+(1)tan(ln(ln(2)))=-0.384

y_2=-0.384+(1)tan(ln(ln(3)))=-0.290

y_3=-0.290+(1)tan(ln(ln(4)))=0.049

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Example Question #745 : Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that f'(x)=sec(x^4) and y(1)=10 , approximate y(2) using Euler's Method and four steps.

Possible Answers:

10.615

10.725

10.422

10.983

10.515

Correct answer:

10.615

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

To calculate the step size find the distance between the final and initial value and divide by the number of steps to be used:

$\Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told f'(x)=sec(x^4) and y(1)=10

Knowing this, we may take the steps to estimate our function value at our final value:

$\Delta x = \frac{2-1}{4}=0.25$

y_0=10;x_0=1

y_1=10+(0.25)sec(1^4)=10.463

y_2=10.463+(0.25)sec(1.25^4)=10.136

y_3=10.136+(0.25)sec(1.5^4)=10.865

y_4=10.865+(0.25)sec(1.75^4)=10.615

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Example Question #746 : Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that f'(x)=csc(2^x) and y(1)=3 , approximate y(5) using Euler's Method and four steps.

Possible Answers:

0.317

4.100

1.208

2.779

3.790

Correct answer:

0.317

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

To calculate the step size find the distance between the final and initial value and divide by the number of steps to be used:

$\Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told f'(x)=csc(2^x) and y(1)=3

Knowing this, we may take the steps to estimate our function value at our final value:

$\Delta x = \frac{5-1}{4}=1$

y_0=3;x_0=1

y_1=3+(1)csc(2^1)=4.100

y_2=4.100+(1)csc(2^2)=2.779

y_3=2.779+(1)csc(2^3)=3.790

y_4=3.790+(1)csc(2^4)=0.317

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Example Question #747 : Differential Functions

The expression of a particular function is unknown; however, we have an expression for its derivative. Knowing that $f'(x)=11^{x^6}$ and y(1)=5 , approximate y(1.16) using Euler's Method and four steps.

Possible Answers:

12.614

15.622

12.025

8.098

13.159

Correct answer:

12.614

Explanation:

The general form of Euler's method, when a derivative function, initial value, and step size are known, is:

$y_n=y_{n+1} +\Delta x f'(x_n,y_n)$

To calculate the step size find the distance between the final and initial value and divide by the number of steps to be used:

$\Delta x = \frac{x_f-x_i}{Steps}$

For this problem, we are told $f'(x)=11^{x^6}$ and y(1)=5

Knowing this, we may take the steps to estimate our function value at our final value:

$\Delta x = \frac{1.16-1}{4}=0.04$

y_0=5;x_0=1

$y_1=5+(0.04)11^{1^6}=5.44$

$y_2=5.44+(0.04)11^{1.04^6}=6.271$

$y_3=6.271+(0.04)11^{1.08^6}=8.068$

$y_4=8.068+(0.04)11^{1.12^6}=12.614$

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #551 : How To Find Differential Functions

Example Question #739 : Differential Functions

Example Question #740 : Differential Functions

Example Question #741 : Differential Functions

Example Question #742 : Differential Functions

Example Question #743 : Differential Functions

Example Question #744 : Differential Functions

Example Question #745 : Differential Functions

Example Question #746 : Differential Functions

Example Question #747 : Differential Functions

All Calculus 1 Resources

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