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Calculus 1 : Differential Functions

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Example Questions

Example Question #431 : Functions

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

Possible Answers:

$8e^{2xy}+16xye^{2xy}+16x^2y^2e^{2xy}$

$8e^{2xy}+8xye^{2xy}+8x^2y^2e^{2xy}$

$4e^{2xy}+32xye^{2xy}+8x^2y^2e^{2xy}$

$8e^{2xy}+32xye^{2xy}+8x^2y^2e^{2xy}$

$8e^{2xy}+32xye^{2xy}+16x^2y^2e^{2xy}$

Correct answer:

$8e^{2xy}+32xye^{2xy}+16x^2y^2e^{2xy}$

Explanation:

For this problem, we're asked to take the derivative of a function multiple times, each time with respect to a particular variable.

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:

For the function

$f(x,y)=e^{2xy}$

We'll make use of the following derivative properties:

Derivative of an exponential: d[e^u]=du(e^u)

Product rule: d[uv]=udv+vdu

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

Looking for $f_{xyyx}$ , begin by taking the derivative with respect to x:

$f_x=2ye^{2xy}$

Take the derivative once more, this time with respect to y:

$f_{xy}=2e^{2xy}+4xye^{2xy}$

Take the derivative with respect to y once more:

$f_{xyy}=4xe^{2xy}+4xe^{2xy}+8x^2ye^{2xy}$

$f_{xyy}=8xe^{2xy}+8x^2ye^{2xy}$

Finally, take the derivative with respect to x:

$f_{xyyx}=8e^{2xy}+16xye^{2xy}+16xye^{2xy}+16x^2y^2e^{2xy}$

$f_{xyyx}=8e^{2xy}+32xye^{2xy}+16x^2y^2e^{2xy}$

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Example Question #241 : How To Find Differential Functions

Find $f_{xy}$ of the function $f(x)=x^2y^2e^{x^2y^2}$ .

Possible Answers:

$4xye^{x^2y^2}+4x^3y^3e^{x^2y^2}+4x^5y^5e^{x^2y^2}$

$4xye^{x^2y^2}+12x^3y^3e^{x^2y^2}+16x^5y^5e^{x^2y^2}$

$4xye^{x^2y^2}+12x^3y^3e^{x^2y^2}+3x^5y^5e^{x^2y^2}$

$4xye^{x^2y^2}+12x^3y^3e^{x^2y^2}+8x^5y^5e^{x^2y^2}$

$4xye^{x^2y^2}+12x^3y^3e^{x^2y^2}+4x^5y^5e^{x^2y^2}$

Correct answer:

$4xye^{x^2y^2}+12x^3y^3e^{x^2y^2}+4x^5y^5e^{x^2y^2}$

Explanation:

For this problem, we're asked to take the derivative of a function multiple times, each time with respect to a particular variable.

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:

For the function

$f(x)=x^2y^2e^{x^2y^2}$

We'll make use of the following derivative properties:

Derivative of an exponential: d[e^u]=du(e^u)

Product rule: d[uv]=udv+vdu

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

Since we're looking for $f_{xy}$ , begin by taking the derivative with respect to x:

$f_x=2xy^2e^{x^2y^2}+2x^3y^4e^{x^2y^2}$

Now, take the derivative with respect to y:

$f_{xy}=4xye^{x^2y^2}+4x^3y^3e^{x^2y^2}+8x^3y^3e^{x^2y^2}+4x^5y^5e^{x^2y^2}$

$f_{xy}=4xye^{x^2y^2}+12x^3y^3e^{x^2y^2}+4x^5y^5e^{x^2y^2}$

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Example Question #241 : How To Find Differential Functions

Which of the following functions is continuous but not differentiable at x=0?

Possible Answers:

$f(x)=x^{2}$

$f(x)=\frac{1}{x}$

$f(x)=\sqrt{x-1}$

$f(x)=\sqrt[16]{x}$

Correct answer:

$f(x)=\sqrt[16]{x}$

Explanation:

Each of these functions gives a different case. $f(x)=x^{2}$ is a polynomial, and is consequently continuous and differentiable for the entire real line. $f(x)=\frac{1}{x}$ is not continuous at x=0 since it is undefined. $f(x)=\sqrt{x-1}$ yields a negative radicand at x=0 and consequently a complex-valued answer (so it is not continuous since we are studying functions of real variables). Finally, $f(x)=\sqrt[16]{x}$ is continuous at x=0 ( f(x)=0 and the limit from the right equals zero, which is all that is necessary because the function is right handed from x=0 ). The derivative of f(x) , however is $\frac{1}{16}x^{-15/16}$ , which is undefined at x=0 , so f(x) is not differentiable at x=0 .

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Example Question #241 : How To Find Differential Functions

Which of the following is an expression for the derivative of $f(x)=x^{2}+1$ ?

Possible Answers:

$\lim_{h->0} \frac{(x^{2}+h^{2})-x^{2}-1}{h}$

$\lim_{h->0} \frac{(x^{2}+2xh+h^{2})-x^{2}+1}{h}$

$\lim_{h->0} \frac{(x^{2}+h^{2})-(x^{2}-1)}{h}$

$\lim_{h->0} \frac{(x^{2}+1)-h^{2}-1}{h}$

Correct answer:

$\lim_{h->0} \frac{(x^{2}+2xh+h^{2})-x^{2}+1}{h}$

Explanation:

This requires remembering the difference quotient and being careful with algebra. Much of the trouble that arises in calculating derivatives from the definition (and down the road, stickier integrals) is in faulty algebra. When plugging in x+h for when calculating f(x+h) , be sure to FOIL. Also, when subtracting a sum of terms in parantheses, be sure to distribute the negative sign.

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Example Question #435 : Functions

Find the gradient of the function

$f(x,y,z)=x^2y+e^{2xz}$

Possible Answers:

$(xy+ze^{2xz})\widehat{i}+x^2\widehat{j}+(x^2y+2xe^{2xz})\widehat{k}$

$(2xy+2ze^{2xz})\widehat{i}+x^2\widehat{j}+2xe^{2xz}\widehat{k}$

$(2xy+2ze^{2xz})\widehat{i}+(x^2+e^{2xz})\widehat{j}+(x^2y+2xe^{2xz})\widehat{k}$

$(2xy)\widehat{i}+x^2\widehat{j}+2xe^{2xz}\widehat{k}$

$(2xy+2ze^{2xz})\widehat{i}+e^{2xz}\widehat{j}+x^2y\widehat{k}$

Correct answer:

$(2xy+2ze^{2xz})\widehat{i}+x^2\widehat{j}+2xe^{2xz}\widehat{k}$

Explanation:

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:

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

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

Knowledge of the following derivative rules will be necessary:

Derivative of an exponential: d[e^u]=e^udu

Note that u and v 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.

The function we're considering is

$f(x,y,z)=x^2y+e^{2xz}$

The derivatives follow as

$\frac{\delta f}{\delta x}= 2xy+2ze^{2xz}$

$\frac{\delta f}{\delta y}=x^2$

$\frac{\delta f}{\delta z}= 2xe^{2xz}$

The gradient is thus the sum of these partials, multiplied by their respective vectors:

$(2xy+2ze^{2xz})\widehat{i}+x^2\widehat{j}+2xe^{2xz}\widehat{k}$

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Example Question #436 : Functions

Find the gradient of the function

$f(x,y)=xy^2e^{3x^2y}$

Possible Answers:

$xye^{3x^2y}(2+3x^2y)\widehat{i}+y^2e^{3x^2y}(1+6x^2y)\widehat{j}$

$y^4e^{3x^2y}(1+6x^2y)\widehat{i}+x^3ye^{3x^2y}(2+3x^2y)\widehat{j}$

$e^{3x^2y}(y^2+6x^2y^3+2xy+3x^3y^2)$

$e^{3x^2y}(1+6x^2y)\widehat{i}+xe^{3x^2y}(2+3x^2y)\widehat{j}$

$y^2e^{3x^2y}(1+6x^2y)\widehat{i}+xye^{3x^2y}(2+3x^2y)\widehat{j}$

Correct answer:

$y^2e^{3x^2y}(1+6x^2y)\widehat{i}+xye^{3x^2y}(2+3x^2y)\widehat{j}$

Explanation:

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 rules will be necessary:

Derivative of an exponential: d[e^u]=e^udu

Product rule: d[uv]=udv+vdu

Note that u and v 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.

For the function

$f(x,y)=xy^2e^{3x^2y}$

$\frac{\delta f}{\delta x} =y^2e^{3x^2y}+6x^2y^3e^{3x^2y}$

$\frac{\delta f}{\delta y} = 2xye^{3x^2y}+3x^3y^2e^{3x^2y}$

The gradient is then the sum of these multiplied by their respective vectors

$y^2e^{3x^2y}(1+6x^2y)\widehat{i}+xye^{3x^2y}(2+3x^2y)\widehat{j}$

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Example Question #241 : Other Differential Functions

What is the derivative of the function $f(x,y,z)=\sin(x)\cos(y)e^z$ ?

Possible Answers:

$e^z(\sin(x)\sin(y)dx-\cos(x)\cos(y)dy+\sin(x)\cos(y)dz)$

$e^z(\sin(x)\cos(y)dx+\sin(x)\sin(y)dy-\cos(x)\cos(y)dz)$

$e^z(\cos(x)\cos(y)dx-\sin(x)\sin(y)dy+\sin(x)\cos(y)dz)$

$e^z(\sin(x)\cos(y)dx-\sin(x)\sin(y)dy+\sin(x)\cos(y)dz)$

Correct answer:

$e^z(\cos(x)\cos(y)dx-\sin(x)\sin(y)dy+\sin(x)\cos(y)dz)$

Explanation:

Note that for this problem, we're not told to take the derivative with respect to any particular variable, so it would be prudent to take the derivative with respect to all three. Knowledge of the following derivative rules will be necessary:

Derivative of an exponential: d[e^u]=e^udu

Trigonometric derivative: $d[\sin(u)]=\cos(u)du;d[\cos(u)]=-\sin(u)du$

Product rule: d[uv]=udv+vdu

Note that and 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.

Looking at our function, take the derivative with respect to each variable

$f(x,y,z)=\sin(x)\cos(y)e^z$

$f_x=\cos(x)\cos(y)e^zdx$

$f_y=-\sin(x)\sin(y)e^zdy$

$f_z=\sin(x)\cos(y)e^zdz$

The complete derivative is then the sum of these:

$f'(x,y,z)=e^z(\cos(x)\cos(y)dx-\sin(x)\sin(y)dy+\sin(x)\cos(y)dz)$

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Example Question #438 : Functions

The magnitude of a particle's noise varies with position and is given by the function

$f(x,y,z)=x^2ye^{xz^2}\sin(y+z)$

Describe rate of the change in magnitude as the value of x changes.

Possible Answers:

$xye^{xz^2}\sin(y+z)(2+xz^2)$

$xye^{xz^2}\cos(y+z)(2+xz^2)$

$xze^{xz^2}\sin(y+z)(2+xy^2)$

$xze^{xz^2}\cos(y+z)(2+xy^2)$

$xy^2e^{xz^2}\sin(y+z)(2+xz^2)$

Correct answer:

$xye^{xz^2}\sin(y+z)(2+xz^2)$

Explanation:

Rates of change of a function can be found by taking the derivative of the function. Note that a rate of change can depend on multiple variables, and sometimes we only wish to know the influence of one variable

For this problem, we're being asked to take the derivative with respect to one particular variable, in this case x. This is known as taking a partial derivative; often it is denoted with the Greek character delta, $\delta$ , or by the subscript of the variable being considered such as f_x or f_y .

Knowledge of the following derivative rules will be necessary:

Derivative of an exponential: d[e^u]=e^udu

Product rule: d[uv]=udv+vdu

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

The function we're given is

$f(x,y,z)=x^2ye^{xz^2}\sin(y+z)$

We'll be taking the derivative with respect to x using the above rules:

$\frac{\delta f(x,y,z)}{\delta x}=2xye^{xz^2}\sin(y+z)+x^2yz^2e^{xz^2}\sin(y+z)$

$\frac{\delta f(x,y,z)}{\delta x}=xye^{xz^2}\sin(y+z)(2+xz^2)$

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Example Question #1461 : Calculus

Take the derivative with respect to y of the function $f(x,y,z)=x^2yz^3\cos(xy^2z)$

Possible Answers:

$x^4z^2(\cos(xy^2z)-2xy^2z\sin(xy^2z))$

$x^3z^3(\cos(xy^2z)-2xyz\sin(xy^2z))$

$x^2z^3(\cos(xy^2z)-2xy^2z\sin(xy^2z))$

$xz(\cos(xy^2z)+xy^2z\sin(xy^2z))$

$x^2z(\cos(xy^2z)-2xy^4z\sin(xy^2z))$

Correct answer:

$x^2z^3(\cos(xy^2z)-2xy^2z\sin(xy^2z))$

Explanation:

Note that for this problem, we're told to take the derivative with respect to one particular variable. This is known as taking a partial derivative; often it is denoted with the Greek character delta, $\delta$ , or by the subscript of the variable being considered such as f_x or f_y .

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative: $d[\cos(u)]=-\sin(u)du$

Product rule: d[uv]=udv+vdu

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

For the function

$f(x,y,z)=x^2yz^3\cos(xy^2z)$

$\frac{\delta f(x,y,z)}{\delta y} = x^2z^3\cos(xy^2z)-2x^3y^2z^4\sin(xy^2z)$

$\frac{\delta f(x,y,z)}{\delta y} = x^2z^3(\cos(xy^2z)-2xy^2z\sin(xy^2z))$

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Example Question #440 : Functions

The brightness of a lightbulb varies with its position on a board, and is given by the function

$f(x,y,z)=x^2y\sin(x+z)$

What is the affect of the variable x on the rate of change of the bulb's brightness?

Possible Answers:

$x^2y^2(2\sin(x+z)+x\cos(x+z))$

$2\sin(x+z)+x\cos(x+z)$

$yz(2\sin(x+z)+x\cos(x+z))$

$xy(2\sin(x+z)+x\cos(x+z))$

$xy^2(2\sin(x+z)+x\cos(x+z))$

Correct answer:

$xy(2\sin(x+z)+x\cos(x+z))$

Explanation:

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative: $d[\sin(u)]=\cos(u)du$

Product rule: d[uv]=udv+vdu

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

Our function is

$f(x,y,z)=x^2y\sin(x+z)$

and we're curious about the influence of x on the rate of its change, so we'll take the derivative with respect to x:

$\frac{\delta f(x,y,z)}{\delta x}=2xy\sin(x+z)+x^2y\cos(x+z)$

$\frac{\delta f(x,y,z)}{\delta x}=xy(2\sin(x+z)+x\cos(x+z))$

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #431 : Functions

Example Question #241 : How To Find Differential Functions

Example Question #241 : How To Find Differential Functions

Example Question #241 : How To Find Differential Functions

Example Question #435 : Functions

Example Question #436 : Functions

Example Question #241 : Other Differential Functions

Example Question #438 : Functions

Example Question #1461 : Calculus

Example Question #440 : Functions

All Calculus 1 Resources

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