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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 $g(x) = \sqrt{x^3+5}$ $\displaystyle g(x) = \sqrt{x^3+5}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Rewrite the square root as the $\small \frac{1}{2}$ $\displaystyle \small \frac{1}{2}$ 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 #552 : How To Find Differential Functions

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #553 : Other Differential Functions

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

Possible Answers:

4.019 $\displaystyle 4.019$

3.912 $\displaystyle 3.912$

3.749 $\displaystyle 3.749$

3.888 $\displaystyle 3.888$

3.996 $\displaystyle 3.996$

Correct answer:

3.912 $\displaystyle 3.912$

Explanation:

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

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

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

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

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

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

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

$\displaystyle y_0=4;x_0=2$

$\displaystyle y_1=4+(0.01)csc(cot(2))=3.977$

$\displaystyle y_2=3.977+(0.01)csc(cot(2.01))=3.955$

$\displaystyle y_3=3.955+(0.01)csc(cot(2.02))=3.933$

$\displaystyle y_4=3.933+(0.01)csc(cot(2.03))=3.912$

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Example Question #554 : How To Find Differential Functions

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

Possible Answers:

0.383 $\displaystyle 0.383$

1.063 $\displaystyle 1.063$

0.915 $\displaystyle 0.915$

1.561 $\displaystyle 1.561$

2.147 $\displaystyle 2.147$

Correct answer:

1.063 $\displaystyle 1.063$

Explanation:

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

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

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

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

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

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

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

$\displaystyle y_0=2,x_0=3$

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

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

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

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Example Question #552 : Other Differential Functions

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

Possible Answers:

4.783 $\displaystyle 4.783$

5.112 $\displaystyle 5.112$

4.292 $\displaystyle 4.292$

5.305 $\displaystyle 5.305$

5.496 $\displaystyle 5.496$

Correct answer:

5.305 $\displaystyle 5.305$

Explanation:

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

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

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

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

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

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

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

$\displaystyle y_0=4,x_0=0$

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

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

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

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Example Question #1772 : Calculus

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

Possible Answers:

5.276 $\displaystyle 5.276$

4.938 $\displaystyle 4.938$

5.191 $\displaystyle 5.191$

5.072 $\displaystyle 5.072$

5.409 $\displaystyle 5.409$

Correct answer:

5.072 $\displaystyle 5.072$

Explanation:

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

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

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

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

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

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

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

$\displaystyle y_0=5,x_0=1$

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

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

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

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Example Question #1773 : Calculus

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

Possible Answers:

-0.265 $\displaystyle -0.265$

0.049 $\displaystyle 0.049$

-0.517 $\displaystyle -0.517$

0.183 $\displaystyle 0.183$

-0.384 $\displaystyle -0.384$

Correct answer:

0.049 $\displaystyle 0.049$

Explanation:

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

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

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

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

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

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

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

$\displaystyle y_0=0;x_0=2$

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

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

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

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Example Question #1774 : Calculus

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

Possible Answers:

10.615 $\displaystyle 10.615$

10.725 $\displaystyle 10.725$

10.515 $\displaystyle 10.515$

10.983 $\displaystyle 10.983$

10.422 $\displaystyle 10.422$

Correct answer:

10.615 $\displaystyle 10.615$

Explanation:

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

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

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

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

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

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

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

$\displaystyle y_0=10;x_0=1$

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

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

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

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

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Example Question #1775 : Calculus

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

Possible Answers:

1.208 $\displaystyle 1.208$

3.790 $\displaystyle 3.790$

4.100 $\displaystyle 4.100$

2.779 $\displaystyle 2.779$

0.317 $\displaystyle 0.317$

Correct answer:

0.317 $\displaystyle 0.317$

Explanation:

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

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

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

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

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

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

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

$\displaystyle y_0=3;x_0=1$

$\displaystyle y_1=3+(1)csc(2^1)=4.100$

$\displaystyle y_2=4.100+(1)csc(2^2)=2.779$

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

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

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Example Question #1776 : Calculus

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

Possible Answers:

12.614 $\displaystyle 12.614$

12.025 $\displaystyle 12.025$

8.098 $\displaystyle 8.098$

15.622 $\displaystyle 15.622$

13.159 $\displaystyle 13.159$

Correct answer:

12.614 $\displaystyle 12.614$

Explanation:

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

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

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

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

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

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

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

$\displaystyle y_0=5;x_0=1$

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

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

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

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

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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 #552 : How To Find Differential Functions

Example Question #553 : Other Differential Functions

Example Question #554 : How To Find Differential Functions

Example Question #552 : Other Differential Functions

Example Question #1772 : Calculus

Example Question #1773 : Calculus

Example Question #1774 : Calculus

Example Question #1775 : Calculus

Example Question #1776 : Calculus

All Calculus 1 Resources

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