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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #451 : Partial Derivatives

Find $\frac{\partial f}{\partial y}$ for $f \left(x,y \right )=\sin \left(x^2y-8y^2x \right )$ .

Possible Answers:

$\cos \left(x^2y-8y^2x \right ) \left(x^2-16yx \right )$

$\cos \left(x^2y-8y^2x \right ) \left(x^2-16x \right )$

$\sin \left(x^2y-8y^2x \right ) \left(x^2-16yx \right )$

$\cos \left(x^2y-8y^2x \right ) \left(x^2-8x \right )$

Correct answer:

$\cos \left(x^2y-8y^2x \right ) \left(x^2-16yx \right )$

Explanation:

When finding partial derivatives of multi-variable functions with respect to a specific variable, all other variables are treated as a constant. By the chain-rule,

$\frac{\partial }{\partial y}\sin \left(f\left(x,y \right ) \right )= \cos \left(f\left(x,y \right ) \right )\cdot f_y \left(x,y \right )$ .

Noting, that

$f \left(x,y \right )= x^2y -8y^2x \Rightarrow f_y = x^2 - 16yx$ .

Plugging all of this in, we arrive at the final result,

$\cos \left(x^2y-8y^2x \right ) \left(x^2-16yx \right )$ .

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Example Question #2821 : Calculus 3

Find $\frac{\partial f}{\partial x}$ of $f \left(x,y \right ) = \ln \left(\frac{x}{y} \right )$

Possible Answers:

$\frac{1}{y}$

$\frac{x}{y}$

$\frac{1}{x}$

$\frac{y}{x}$

Correct answer:

$\frac{1}{x}$

Explanation:

In finding the partial derivative of a multi-variable function with respect to a particular variable, all other variables are treated as constant. We can begin by rewriting our function in the following way

$\ln \left(\frac{x}{y} \right )= \ln x - \ln y$ .

Therefore, when we take a derivative with respect to , we find

$f_x = \frac{1}{x}$ .

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Example Question #455 : Partial Derivatives

Find $\frac{\partial ^2 f}{\partial x \partial y}$ of $f \left(x,y \right )= 3xy^2+4x^2y$

Possible Answers:

6y+8x

3y^2+8xy

Correct answer:

6y+8x

Explanation:

When taking mixed-partials of a multi-variable function, such as $f \left(x,y \right )$ , the order in which the partials derivatives are taken does not matter. Consider the following

$\frac{\partial ^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y} \right )= \frac{\partial }{\partial x} \left(6xy+4x^2 \right )= 6y+8x$ .

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Example Question #456 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xyz}\\&\text{Where }f(x,y,z)=-\frac{ {(4x^3e^{(y^2)})}}{z^2} - 9x^3cos{(y^4)}ln{(z^4)}\end{align*}$

Possible Answers:

$\frac{{(432x^2y^3sin{(y^4)})}}{z} +\frac{ {(48x^2ye^{(y^2)})}}{z^3}$

${\frac{{(8x^3e^{(y^2)})}}{z^3} -\frac{ {(12x^2e^{(y^2)})}}{z^2} -\frac{ {(36x^3cos{(y^4)})}}{z} - 27x^2cos{(y^4)}ln{(z^4)} + 36x^3y^3ln{(z^4)}sin{(y^4)} -\frac{ {(8x^3ye^{(y^2)})}}{z^2} }$

${\frac{{(432x^2y^3sin{(y^4)})}}{z} - 27x^2cos{(y^4)}ln{(z^4)} -\frac{ {(12x^2e^{(y^2)})}}{z^2} + 108x^2y^3ln{(z^4)}sin{(y^4)} -\frac{ {(24x^2ye^{(y^2)})}}{z^2} +\frac{ {(48x^2ye^{(y^2)})}}{z^3} }$

$\frac{{(432x^2y^3sin{(y^4)})}}{z} -\frac{ {(48x^2ye^{(y^2)})}}{z^3}$

Correct answer:

$\frac{{(432x^2y^3sin{(y^4)})}}{z} +\frac{ {(48x^2ye^{(y^2)})}}{z^3}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&d[uv]=udv+vdu\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&f_{x}=-\frac{ {(12x^2e^{(y^2)})}}{z^2} - 27x^2cos{(y^4)}ln{(z^4)}\\&f_{xy}=108x^2y^3ln{(z^4)}sin{(y^4)} -\frac{ {(24x^2ye^{(y^2)})}}{z^2}\\&f_{xyz}=\frac{{(432x^2y^3sin{(y^4)})}}{z} +\frac{ {(48x^2ye^{(y^2)})}}{z^3}\end{align*}$

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Example Question #457 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zx}\\&\text{Where }f(x,y,z)=9\cdot2^{(x^3)}sin{(y^3)}e^{(z)}\end{align*}$

Possible Answers:

$168.4\cdot2^{(2x^3)}x^2sin{(y^3)}^2e^{(2z)}$

${9\cdot2^{(x^3)}sin{(y^3)}e^{(z)} + 18.71\cdot2^{(x^3)}x^2sin{(y^3)}e^{(z)}}$

${18.71\cdot2^{(3x^2)}x^2sin{(3y^2)}e^{(z)}}$

${18.71\cdot2^{(x^3)}x^2sin{(y^3)}e^{(z)}}$

Correct answer:

${18.71\cdot2^{(x^3)}x^2sin{(y^3)}e^{(z)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&d[uv]=udv+vdu\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&f_{z}=9\cdot2^{(x^3)}sin{(y^3)}e^{(z)}\\&f_{zx}=18.71\cdot2^{(x^3)}x^2sin{(y^3)}e^{(z)}\end{align*}$

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Example Question #451 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yxz}\\&\text{Where }f(x,y,z)=-3\cdot4^{(y^4)}sin{(z^3)}ln{(x)}\end{align*}$

Possible Answers:

${-\frac{ {(3\cdot4^{(y^4)}sin{(z^3)})}}{x} - 9\cdot4^{(y^4)}z^2cos{(z^3)}ln{(x)} - 16.64\cdot4^{(y^4)}y^3sin{(z^3)}ln{(x)}}$

${\frac{-{(49.91\cdot4^{(4y^3)}y^3z^2cos{(3z^2)})}}{x}}$

${\frac{-{(49.91\cdot4^{(y^4)}y^3z^2cos{(z^3)})}}{x}}$

$\frac{-{(449.2\cdot4^{(3y^4)}y^3z^2cos{(z^3)}sin{(z^3)}^2ln{(x)}^2)}}{x}$

Correct answer:

${\frac{-{(49.91\cdot4^{(y^4)}y^3z^2cos{(z^3)})}}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&d[a^u]=a^uduln(a)\\&d[ln(u)]=\frac{du}{u}\\&d[sin(u)]=cos(u)du\\&f_{y}=-16.64\cdot4^{(y^4)}y^3sin{(z^3)}ln{(x)}\\&f_{yx}=\frac{-{(16.64\cdot4^{(y^4)}y^3sin{(z^3)})}}{x}\\&f_{yxz}=\frac{-{(49.91\cdot4^{(y^4)}y^3z^2cos{(z^3)})}}{x}\end{align*}$

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Example Question #452 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zx}\\&\text{Where }f(x,y,z)=\frac{-{(6e^{(x^3)}cos{(y)})}}{z}\end{align*}$

Possible Answers:

${\frac{{(6e^{(x^3)}cos{(y)})}}{z^2} +\frac{ {(18x^2e^{(x^3)}cos{(y)})}}{z^2}}$

${\frac{{(6e^{(x^3)}cos{(y)})}}{z^2} -\frac{ {(18x^2e^{(x^3)}cos{(y)})}}{z}}$

${\frac{{(18x^2e^{(x^3)}cos{(y)})}}{z^2}}$

$\frac{-{(108x^2e^{(2x^3)}cos{(y)}^2)}}{z^3}$

Correct answer:

${\frac{{(18x^2e^{(x^3)}cos{(y)})}}{z^2}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&d[e^u]=e^udu\\&f_{z}=\frac{{(6e^{(x^3)}cos{(y)})}}{z^2}\\&f_{zx}=\frac{{(18x^2e^{(x^3)}cos{(y)})}}{z^2}\end{align*}$

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Example Question #455 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yzy}\\&\text{Where }f(x,y,z)=10\cdot4^{(z^2)}xsin{(y^3)}\end{align*}$

Possible Answers:

${27.73\cdot4^{(z^2)}xzsin{(y^3)} + 60\cdot4^{(z^2)}xy^2cos{(y^3)}}$

$24955\cdot4^{(3z^2)}x^3y^4zcos{(y^3)}^2sin{(y^3)}$

${166.4\cdot4^{(z^2)}xyzcos{(y^3)} - 249.5\cdot4^{(z^2)}xy^4zsin{(y^3)}}$

${30\cdot4^{(z^2)}xy^2cos{(y^3)} + 166.4\cdot4^{(z^2)}xyzcos{(y^3)} + 83.18\cdot4^{(z^2)}xy^2zcos{(y^3)} - 249.5\cdot4^{(z^2)}xy^4zsin{(y^3)}}$

Correct answer:

${166.4\cdot4^{(z^2)}xyzcos{(y^3)} - 249.5\cdot4^{(z^2)}xy^4zsin{(y^3)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&f_{y}=30\cdot4^{(z^2)}xy^2cos{(y^3)}\\&f_{yz}=83.18\cdot4^{(z^2)}xy^2zcos{(y^3)}\\&f_{yzy}=166.4\cdot4^{(z^2)}xyzcos{(y^3)} - 249.5\cdot4^{(z^2)}xy^4zsin{(y^3)}\end{align*}$

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Example Question #461 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yxx}\\&\text{Where }f(x,y,z)=-5ln{(y^4)}e^{(z)}sin{(x)}\end{align*}$

Possible Answers:

${-\frac{ {(20e^{(z)}sin{(x)})}}{y} - 10ln{(y^4)}e^{(z)}cos{(x)}}$

${\frac{-{(20e^{(z)}cos{(x)})}}{y}}$

${\frac{{(20e^{(z)}sin{(x)})}}{y}}$

$\frac{-{(500ln{(y^4)}^2e^{(3z)}cos{(x)}^2sin{(x)})}}{y}$

Correct answer:

${\frac{{(20e^{(z)}sin{(x)})}}{y}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&d[ln(u)]=\frac{du}{u}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&f_{y}=\frac{-{(20e^{(z)}sin{(x)})}}{y}\\&f_{yx}=\frac{-{(20e^{(z)}cos{(x)})}}{y}\\&f_{yxx}=\frac{{(20e^{(z)}sin{(x)})}}{y}\end{align*}$

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Example Question #2821 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yy}\\&\text{Where }f(x,y,z)=-6cos{(z^2)}ln{(y^2)}\end{align*}$

Possible Answers:

${\frac{-{(24cos{(z^2)})}}{y}}$

$\frac{{(144cos{(z^2)}^2)}}{y^2}$

${\frac{{(12cos{(z^2)})}}{y^2}}$

${\frac{{(12cos{(z^2)})}}{y^2} -\frac{ {(12cos{(z^2)})}}{y}}$

Correct answer:

${\frac{{(12cos{(z^2)})}}{y^2}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&d[ln(u)]=\frac{du}{u}\\&f_{y}=\frac{-{(12cos{(z^2)})}}{y}\\&f_{yy}=\frac{{(12cos{(z^2)})}}{y^2}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #451 : Partial Derivatives

Example Question #2821 : Calculus 3

Example Question #455 : Partial Derivatives

Example Question #456 : Partial Derivatives

Example Question #457 : Partial Derivatives

Example Question #451 : Partial Derivatives

Example Question #452 : Partial Derivatives

Example Question #455 : Partial Derivatives

Example Question #461 : Partial Derivatives

Example Question #2821 : Calculus 3

All Calculus 3 Resources

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