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Calculus 1 : Differential Equations

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Example Questions

Example Question #2431 : Calculus

Find $f_{xyz}$ if $f(x,y,z)=e^{yz}cos(x^2z)$

Possible Answers:

$-2xz^2e^{yz}sin(x^2z)$

$-2xze^{yz}sin(x^2z)$

$-2x(2z^2e^{yz}sin(x^2z)+y^3ze^{yz}sin(x^2z)))$

$-2xz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

$-2yz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

Correct answer:

$-2xz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

Explanation:

For this problem, note that:

d[e^u]=du(e^u)

d[sin(u)]=du(cos(u))

d[cos(u)]=-du(sin(u))

Product Rule d[uv]=udv+vdu

To solve this problem, differentiate the expression one variable at a time, treating other variables as constants:

If we're looking for $f_{xyz}$ for the function $f(x,y,z)=e^{yz}cos(x^2z)$ then we'll begin by differentiating with respect to first:

$f_x=-2xze^{yz}sin(x^2z)$

Next, differentiate with respect to :

$f_{xy}=-2xz^2e^{yz}sin(x^2z)$

Now finally we'll differentiate with respect to ; remember to use the product rule:

$f_{xyz}=-4xze^{yz}sin(x^2z)-2xz^2(ye^{yz}sin(x^2z)+x^2e^{yz}cos(x^2z))$

$f_{xyz}=-4xze^{yz}sin(x^2z)-2xyz^2e^{yz}sin(x^2z)-2x^3z^2e^{yz}cos(x^2z))$

$f_{xyz}=-2xz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

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Example Question #351 : Equations

Find $f_{xyy}$ for the function $f(x,y)=x^2e^{y^2}$

Possible Answers:

$12xy^2e^{y^2}$

$4xe^{y^2}+4xy^2e^{y^2}$

$4xe^{y^2}+8xy^2e^{y^2}$

$4ye^{y^2}$

$12xe^{y^2}$

Correct answer:

$4xe^{y^2}+8xy^2e^{y^2}$

Explanation:

For this problem, note that:

d[e^u]=du(e^u)

Product Rule: d[uv]=udv+vdu

To solve this problem, differentiate the expression one variable at a time, treating other variables as constants:

To find $f_{xyy}$ for the function $f(x,y)=x^2e^{y^2}$ , begin by differentiating with respect to :

$f_{x}=2xe^{y^2}$

Next, differentiate with respect to :

$f_{xy}=4xye^{y^2}$

Finally, differentiate with respect to once more, remembering to utilize the product rule:

$f_{xyy}=4xe^{y^2}+8xy^2e^{y^2}$

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Example Question #2433 : Calculus

Find $\frac{dy}{dx}$ for the equation

sin(x^2y)=cos(xy^2)

Possible Answers:

$\frac{2xycos(x^2y)+y^2sin(xy^2)}{2xysin(xy^2)+x^2cos(x^2y)}$

$\frac{2xysin(xy^2)+x^2cos(x^2y)}{2xycos(x^2y)+y^2sin(xy^2)}$

$\frac{-xysin(xy^2)-x^2cos(x^2y)}{xycos(x^2y)+y^2sin(xy^2)}$

$\frac{-2xysin(xy^2)-x^2cos(x^2y)}{2xycos(x^2y)+y^2sin(xy^2)}$

$\frac{2xycos(x^2y)+y^2sin(xy^2)}{-2xysin(xy^2)-x^2cos(x^2y)}$

Correct answer:

$\frac{2xycos(x^2y)+y^2sin(xy^2)}{-2xysin(xy^2)-x^2cos(x^2y)}$

Explanation:

For this problem, note that:

d[sin(u)]=du(cos(u))

d[cos(u)]=-du(sin(u))

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

sin(x^2y)=cos(xy^2)

The derivative is then

2xycos(x^2y)dx+x^2cos(x^2y)dy=-y^2sin(xy^2)dx-2xysin(xy^2)dy

Now bring and terms to opposite sides of the equation:

(2xycos(x^2y)+y^2sin(xy^2))dx=(-2xysin(xy^2)-x^2cos(x^2y))dy

Finally, rearrange terms to find $\frac{dy}{dx}$ :

$\frac{2xycos(x^2y)+y^2sin(xy^2)}{-2xysin(xy^2)-x^2cos(x^2y)}=\frac{dy}{dx}$

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Example Question #2434 : Calculus

Find the derivative of the following function.

$f(x)=\sin (3x)e^{4x}$

Possible Answers:

$12e^{x}\cos(x)$

$3e^{4x}\cos(3x)$

$12e^{4x}\cos(3x)$

None of these

$4e^{4x}\cos(3x)$

Correct answer:

$12e^{4x}\cos(3x)$

Explanation:

To solve this derivative, we must realize that there are two parts to the function and we must use the product rule. The rule states that [f(x)g(x)]'=f'(x)g(x)+f(x)g'(x) .

We must asle recognize that the derivative of $\sin$ is $\cos$ and the derivative of e^x is e^x . By these rules, the derivative is

$f'(x)=3\cos (3x)\cdot4e^{4x}$

$=12e^{4x}\cos(3x)$

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Example Question #1401 : Functions

Find the derivative of the function.

$f(x)=\ln(e^{15x})$

Possible Answers:

None of these

$e^{15x}$

$\frac{1}{15x}$

Correct answer:

Explanation:

This function is just a function inside of a function. This means we have to use the chain rule. The chain rule states that the derivative of [f(g(x))]'=f'(g(x))g'(x) .

The derivative of $\ln(x)$ is $\frac{1}{x}$ and the derivative of e^x is e^x . This makes the derivative

$f'(x)=\frac{1}{e^{15x}}*15e^{15x}$

$=15\frac{e^{15x}}{e^{15x}}$

=15

This makes sense because ln(e^x)=x .

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Example Question #2436 : Calculus

Find the derivative of the function.

$f(x)=\frac{\cos (x)}{x^3}$

Possible Answers:

$-\frac{x\sin (x)+\cos (x)}{x^3}$

$\frac{x\sin (x)+2\cos (x)}{x^3}$

None of these

$-\frac{x\sin (x)+2\cos (x)}{x^3}$

$-\frac{x\sin (x)+2\cos (x)}{x^4}$

Correct answer:

$-\frac{x\sin (x)+2\cos (x)}{x^3}$

Explanation:

To find the derivative of this function, we must use the division rule. This rule states that the derivative of $\frac{f(x)}{g(x)}$ is $\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$ . The derivative of $\cos$ is $-\sin$ and the derivative of x^n is $nx^{n-1}$ .

Thus the derivative is

$f'(x)=\frac{-x^2\sin (x)-2x\cos (x)}{x^4}$

$=-\frac{x\sin (x)+2\cos (x)}{x^3}$

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Example Question #1411 : Functions

Find the derivative of the function.

$f(x)=x\sin (\cos(x))$

Possible Answers:

$x\cos (\cos (x))\cdot\sin (x)+\sin (\cos (x))$

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

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

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

None of these

Correct answer:

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

Explanation:

To find the derivative of this function we need to use the chain rule and multiplication rule. The chain rule states that the derivative of f(g(x)) is f'(g(x))g'(x) . The multiplication rule states that the derivative of f(x)g(x) is f'(x)g(x)+f(x)g'(x) . The derivative of x^n is $nx^{n-1}$ . The derivative of sin is cos and the derivative of cos is -sin. So lets say

g(x)=x then g'(x)=1 and

$h(x)=\sin (\cos (x))$ then $h'(x)=-\cos(\cos (x))\cdot\sin (x)$

So the answer is

$f'(x)=-x\cos (\cos (x))\cdot\sin (x)+\sin (\cos (x))$

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Example Question #1412 : Functions

What is the slope of the function f(x,y)=sin(sin(xy^2)) at the point $(\pi,0)$ ?

Possible Answers:

$(\pi^2,0)$

$(\pi,0)$

$(0,\pi^2)$

$(0,\pi)$

(0,0)

Correct answer:

$(\pi^2,0)$

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of f(x,y)=sin(sin(xy^2)) at the point $(\pi,0)$

$\frac{\delta f}{\delta x}=y^2cos(xy^2)cos(sin(xy^2));\pi^2cos((0)\pi^2)cos(sin((0)\pi^2))=\pi^2$

$\frac{\delta f}{\delta y}=2xycos(xy^2)cos(sin(xy^2));2(0)(\pi)cos((0)\pi^2)cos(sin((0)\pi^2))=0$

The slope is $(\pi^2,0)$

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Example Question #102 : How To Find Solutions To Differential Equations

Find the slope of the function $f(x,y)=x(\frac{1}{2}^{4y})$ at the point (4,1) .

Possible Answers:

(-0.0625,-0.6931)

(0.0625,-0.6931)

(0.0625,0.6931)

(0.6931,0.0625)

(0.6931,-0.0625)

Correct answer:

(0.0625,-0.6931)

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Derivative of an exponential:

d[a^u]=a^uduln(a)

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of $f(x,y)=x(\frac{1}{2}^{4y})$ at the point (4,1)

$\frac{\delta f}{\delta x}=\frac{1}{2}^{4y};\frac{1}{2}^{4(1)}=0.0625$

$\frac{\delta f}{\delta y}=4x(\frac{1}{2}^{4y})ln(\frac{1}{2});4(4)(\frac{1}{2}^{4(1)})ln(\frac{1}{2})=-0.6931$

The slope is (0.0625,-0.6931)

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Example Question #1412 : Functions

Find the slope of the function $f(x,y)=\frac{1}{4}^{\frac{1}{3}xy}$ at the point (2,4) .

Possible Answers:

(0.023,0.046)

(0.046,0.023)

(-0.023,-0.046)

(-0.046,-0.023)

(-0.046,0.023)

Correct answer:

(-0.046,-0.023)

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Derivative of an exponential:

d[a^u]=a^uduln(a)

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of $f(x,y)=\frac{1}{4}^{\frac{1}{3}xy}$ at the point (2,4)

$\frac{\delta f}{\delta x}=\frac{y}{3}(\frac{1}{4}^{\frac{1}{3}xy})ln(\frac{1}{4});\frac{4}{3}(\frac{1}{4}^{\frac{1}{3}(2)(4)})ln(\frac{1}{4})=-0.046$

$\frac{\delta f}{\delta y}=\frac{x}{3}(\frac{1}{4}^{\frac{1}{3}xy})ln(\frac{1}{4});\frac{2}{3}(\frac{1}{4}^{\frac{1}{3}(2)(4)})ln(\frac{1}{4})=-0.023$

The slope is (-0.046,-0.023) .

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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2431 : Calculus

Example Question #351 : Equations

Example Question #2433 : Calculus

Example Question #2434 : Calculus

Example Question #1401 : Functions

Example Question #2436 : Calculus

Example Question #1411 : Functions

Example Question #1412 : Functions

Example Question #102 : How To Find Solutions To Differential Equations

Example Question #1412 : Functions

All Calculus 1 Resources

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