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Calculus 1 : Differential Functions

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Example Questions

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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Example Question #731 : Differential Functions

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

Possible Answers:

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

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

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

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

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 #552 : Other Differential Functions

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

Possible Answers:

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

$\frac{1}{2x}$

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

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

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

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 #553 : Other 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

3.912

3.749

3.888

3.996

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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Calculus 1 : 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 #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

Example Question #731 : Differential Functions

Example Question #552 : Other Differential Functions

Example Question #553 : Other Differential Functions

All Calculus 1 Resources

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