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Calculus 1 : Other Differential Functions

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Example Questions

Example Question #541 : Other Differential Functions

Given that $\frac{dy}{dx}=\ln(x^3)$ and y(1)=13 , use Euler's method to approximate y(1.9) using three steps.

Possible Answers:

15.44

13.66

14.12

14.89

Correct answer:

13.66

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.9-1}{3}=0.3$

Now, to approximate function values using Euler's method, utilize the following formula:

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

After that, it's merely a matter of taking the steps:

y_0=13,x_0=1

$y_1=13+0.3(\ln(1^3))=13$

$y_2=13+0.3(\ln(1.3^3))=13.24$

$y_3=13.24+0.3(\ln(1.6^3))=13.66$

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Example Question #542 : Other Differential Functions

Given that $\frac{dy}{dx}=2^{\ln(x^4)}$ and y(1)=3 , use Euler's method to approximate y(1.8) using four steps.

Possible Answers:

4.04

4.93

5.22

4.46

4.78

Correct answer:

4.78

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.8-1}{4}=0.2$

Now, to approximate function values using Euler's method, utilize the following formula:

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

After that, it's merely a matter of taking the steps:

y_0=3;x_0=1

$y_1=3+0.2(2^{\ln(1^4)})=3.2$

$y_2=3.2+0.2(2^{\ln(1.2^4)})=3.53$

$y_3=3.53+0.2(2^{\ln(1.4^4)})=4.04$

$y_4=4.04+0.2(2^{\ln(1.6^4)})=4.78$

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Example Question #543 : Other Differential Functions

Given that $\frac{dy}{dx}=\ln(2^{x})$ and y(0)=6 , use Euler's method to approximate y(1.5) using three steps.

Possible Answers:

6.520

6.173

6.782

6.999

6.347

Correct answer:

6.520

Explanation:

When using Euler's method, the first step is to calculate step size:

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

Now, to approximate function values using Euler's method, utilize the following formula:

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

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=\ln(2^{x}),y_0=6,x_0=0$

$y_1=6+0.5(\ln(2^{0}))=6$

$y_2=6+0.5(\ln(2^{0.5}))=6.173$

$y_3=6.173+0.5(\ln(2^{1}))=6.520$

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Example Question #541 : Other Differential Functions

Given that $\frac{dy}{dx}=\sin(x^2)$ and y(0)=0 , use Euler's method to approximate y(2) using five steps.

Possible Answers:

0.219

1.424

0.700

0.763

0.920

Correct answer:

0.920

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{2-0}{5}=0.4$

Now, to approximate function values using Euler's method, utilize the following formula:

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

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=\sin(x^2);y_0=0;x_0=0$

$y_1=0+0.4\sin(0^2)=0$

$y_2=0+0.4\sin(0.4^2)=0.064$

$y_3=0.064+0.4\sin(0.8^2)=0.303$

$y_4=0.303+0.4\sin(1.2^2)=0.700$

$y_5=0.700+0.4\sin(1.6^2)=0.920$

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Example Question #542 : Other Differential Functions

Given that $\frac{dy}{dx}= \sin(\sin(\tan(\cos(x))))$ and y(2)=2 , use Euler's method to approximate y(2.3) using three steps.

Possible Answers:

1.756

1.851

1.803

1.942

1.909

Correct answer:

1.851

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{2.3-2}{3}=0.1$

Now, to approximate function values using Euler's method, utilize the following formula:

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

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=\sin(\sin(\tan(\cos(x))));y_0=2;x_0=2$

$y_1=2+0.1(\sin(\sin(\tan(\cos(2)))))=1.959$

$y_2=1.959+0.1(\sin(\sin(\tan( \cos(2.1)))))=1.909$

$y_3=1.909+0.1(\sin(\sin( \tan(\cos(2.2)))))=1.851$

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Example Question #543 : Other Differential Functions

Given that $\frac{dy}{dx}=\cot(x^4)$ and y(1)=5 , use Euler's method to approximate y(1.9) using three steps.

Possible Answers:

4.686

5.799

6.082

4.211

5.253

Correct answer:

5.253

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.9-1}{3}=0.3$

Now, to approximate function values using Euler's method, utilize the following formula:

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

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=\cot(x^4)$

y_0=5;x_0=1

$y_1=5+0.3\cot(1^4)=5.193$

$y_2=5.193+0.3\cot(1.3^4)=4.171$

$y_3=4.171+0.3\cot(1.6^4)=5.253$

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Example Question #544 : Other Differential Functions

Given that $\frac{dy}{dx}=\ln(\sin(\cos(x)))$ and y(0)=4 , use Euler's method to approximate y(1.2) using three steps.

Possible Answers:

3.396

4.012

3.823

3.556

3.663

Correct answer:

3.663

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.2-0}{3}=0.4$

Now, to approximate function values using Euler's method, utilize the following formula:

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

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=\ln(\sin(\cos(x)))$

y_0=4;x_0=0

$y_1=4+0.4\ln(\sin(\cos(0)))=3.931$

$y_2=3.931+0.4\ln(\sin(\cos(0.4)))=3.840$

$y_3=3.840+0.4\ln(\sin(\cos(0.8)))=3.663$

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Example Question #545 : Other Differential Functions

Given that $\frac{dy}{dx}=\frac{\tan(x)}{x^3}$ and y(0.4)=2 , use Euler's method to approximate y(0.44) using four steps.

Possible Answers:

2.022

2.247

2.879

2.356

1.190

Correct answer:

2.247

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{0.44-0.4}{4}=0.01$

Now, to approximate function values using Euler's method, utilize the following formula:

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

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=\frac{\tan(x)}{x^3}$

y_0=2;x_0=0.4

$y_1=2+0.01\frac{\tan(0.4)}{0.4^3}=2.066$

$y_2=2.066+0.01\frac{\tan(0.41)}{0.41^3}=2.129$

$y_3=2.129+0.01\frac{\tan(0.42)}{0.42^3}=2.189$

$y_4=2.189+0.01\frac{\tan(0.43)}{0.43^3}=2.247$

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Example Question #541 : How To Find Differential Functions

Find the derivative of the function $f(x) = e^{(3x+5)}$

Possible Answers:

3e^5

3x+5

$15e^{3x}$

$e^{(3x+5)}$

$3e^{(3x+5)}$

Correct answer:

$3e^{(3x+5)}$

Explanation:

The derivative of $\small e^x$ is $\small e^x$ . You must use either substitution or the chain rule to evaluate more complicated expressions involving $\small e$ . Here, the chain rule is the most efficient.

The derivative of $\small 3x + 5$ is simply 3. So the "inner function" has derivative 3, and the "outer function" is an exponential, which remains the same when differentiated. Therefore, the answer is the original function multiplied by 3.

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Example Question #542 : How To Find Differential Functions

Find the derivative of $\small h(x) = \sin(x) + \cos(x)$

Possible Answers:

$\small -\cos(x) + \sin(x)$

$\small \tan(x)$

$\small \cos(x) - \sin(x)$

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

Correct answer:

$\small \cos(x) - \sin(x)$

Explanation:

The derivative of $\small \sin(x)$ is $\small \cos(x)$ .

The derivative of $\small \cos(x)$ is $\small -\sin(x)$ .

The derivative of the sum is the sum of the derivatives - so we can take the derivatives separately and add them.

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Calculus 1 : Other Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #541 : Other Differential Functions

Example Question #542 : Other Differential Functions

Example Question #543 : Other Differential Functions

Example Question #541 : Other Differential Functions

Example Question #542 : Other Differential Functions

Example Question #543 : Other Differential Functions

Example Question #544 : Other Differential Functions

Example Question #545 : Other Differential Functions

Example Question #541 : How To Find Differential Functions

Example Question #542 : How To Find Differential Functions

All Calculus 1 Resources

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