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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 #783 : Partial Derivatives

Find $f_{zx}$ for the following function:

$f(x, y, z)=\frac{z}{y}+z^4y^3+e^{cos(xy)}z$

Possible Answers:

$\sin(xy)ye^{\cos(xy)}$

$e^{\cos(xy)}$

$\frac{1}{y}+4x^3y^3+e^{\cos(xy)}$

$-\sin(xy)ye^{\cos(xy)}$

Correct answer:

$-\sin(xy)ye^{\cos(xy)}$

Explanation:

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

To start, we must find the partial derivative of the function with respect to z:

$f_z=\frac{1}{y}+4z^3y^3+e^{\cos(xy)}$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Finally, find the partial derivative of the above function with respect to x:

$f_{zx}=-\sin(xy)ye^{\cos(xy)}$

The derivative was found using rules above as well as the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #784 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\\&\text{Where }f(x,y,z)=\frac{(y^{2}tan(11x + x^{2}))}{(3z)}- 2cos(z^{2})sin(x^{3})e^{(y^{3})}\end{align*}$

Possible Answers:

${(\frac{(2ytan(11x + x^{2}))}{(3z)}- 6y^{2}cos(z^{2})sin(x^{3})e^{(y^{3})})\cdot (\frac{(y^{2}\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 6x^{2}cos(x^{3})cos(z^{2})e^{(y^{3})})}$

${\frac{(2y\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 18x^{2}y^{2}cos(x^{3})cos(z^{2})e^{(y^{3})}}$

${(\frac{(2y\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 18x^{2}y^{2}cos(x^{3})cos(z^{2})e^{(y^{3})})\cdot (\frac{(y^{2}\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 6x^{2}cos(x^{3})cos(z^{2})e^{(y^{3})})}$

${\frac{(2ytan(11x + x^{2}))}{(3z)}+\frac{ (y^{2}\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 6x^{2}cos(x^{3})cos(z^{2})e^{(y^{3})} - 6y^{2}cos(z^{2})sin(x^{3})e^{(y^{3})}}$

Correct answer:

${\frac{(2y\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 18x^{2}y^{2}cos(x^{3})cos(z^{2})e^{(y^{3})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(y^{2}tan(11x + x^{2}))}{(3z)}- 2cos(z^{2})sin(x^{3})e^{(y^{3})}\\&f_{x}=\frac{(y^{2}\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 6x^{2}cos(x^{3})cos(z^{2})e^{(y^{3})}\\&f_{xy}=\frac{(2y\cdot (2x + 11)\cdot (tan(11x + x^{2})^{2} + 1))}{(3z)}- 18x^{2}y^{2}cos(x^{3})cos(z^{2})e^{(y^{3})}\end{align*}$

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Example Question #785 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zx}\\&\text{Where }f(x,y,z)=\frac{(2ln(3z)tan(y^{4})e^{(x^{2})})}{5}\end{align*}$

Possible Answers:

${\frac{(8xln(3z)tan(y^{4})^{2}e^{(2x^{2})})}{(25z)}}$

${\frac{(2tan(y^{4})e^{(x^{2})})}{(5z)}+\frac{ (4xln(3z)tan(y^{4})e^{(x^{2})})}{5}}$

${\frac{(4xtan(y^{4})e^{(x^{2})})}{(5z)}}$

${\frac{(8xtan(y^{4})^{2}e^{(2x^{2})})}{(25z^{2})}}$

Correct answer:

${\frac{(4xtan(y^{4})e^{(x^{2})})}{(5z)}}$

Explanation:

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Example Question #786 : Partial Derivatives

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

Possible Answers:

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

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

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

${\frac{(8ln(4z)^{2}e^{(9x)}cos(y^{2} - 4y)sin(y^{2} - 4y)^{2}\cdot (2y - 4)^{2})}{z}}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=2ln(4z)e^{(3x)}cos(y^{2} - 4y)\\&f_{y}=-2ln(4z)e^{(3x)}sin(y^{2} - 4y)\cdot (2y - 4)\\&f_{yy}=- 4ln(4z)e^{(3x)}sin(y^{2} - 4y) - 2ln(4z)e^{(3x)}cos(y^{2} - 4y)\cdot (2y - 4)^{2}\\&f_{yyz}=-\frac{ (4e^{(3x)}sin(y^{2} - 4y))}{z}-\frac{ (2e^{(3x)}cos(y^{2} - 4y)\cdot (2y - 4)^{2})}{z}\end{align*}$

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Example Question #787 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yxy}\\&\text{Where }f(x,y,z)=\frac{(3^yln(4z)tan(x^{4}))}{4}\end{align*}$

Possible Answers:

${\frac{(3^yln(4z)tan(x^{4})ln(3))}{2}+ 3^yx^{3}ln(4z)\cdot (tan(x^{4})^{2} + 1)}$

${3^yx^{3}ln(4z)ln(3)^{2}\cdot (tan(x^{4})^{2} + 1)}$

${\frac{(3^{(3y)}x^{3}ln(4z)^{3}tan(x^{4})^{2}ln(3)^{2}\cdot (tan(x^{4})^{2} + 1))}{16}}$

${\frac{(3^{(3y)}x^{6}ln(4z)^{3}tan(x^{4})ln(3)^{4}\cdot (tan(x^{4})^{2} + 1)^{2})}{4}}$

Correct answer:

${3^yx^{3}ln(4z)ln(3)^{2}\cdot (tan(x^{4})^{2} + 1)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3^yln(4z)tan(x^{4}))}{4}\\&f_{y}=\frac{(3^yln(4z)tan(x^{4})ln(3))}{4}\\&f_{yx}=3^yx^{3}ln(4z)ln(3)\cdot (tan(x^{4})^{2} + 1)\\&f_{yxy}=3^yx^{3}ln(4z)ln(3)^{2}\cdot (tan(x^{4})^{2} + 1)\end{align*}$

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Example Question #788 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yzy}\\&\text{Where }f(x,y,z)=\frac{(5z^{4}cos(y))}{x}-\frac{ (2\cdot 2^{(4z)}sin(y^{2})ln(x))}{9}\end{align*}$

Possible Answers:

${-(\frac{(5z^{4}sin(y))}{x}+\frac{ (4\cdot 2^{(4z)}ycos(y^{2})ln(x))}{9})\cdot (\frac{(20z^{3}sin(y))}{x}+\frac{ (16\cdot 2^{(4z)}ycos(y^{2})ln(2)ln(x))}{9})\cdot (\frac{(20z^{3}cos(y))}{x}+\frac{ (16\cdot 2^{(4z)}cos(y^{2})ln(2)ln(x))}{9}-\frac{ (32\cdot 2^{(4z)}y^{2}sin(y^{2})ln(2)ln(x))}{9})}$

${\frac{(32\cdot 2^{(4z)}y^{2}sin(y^{2})ln(2)ln(x))}{9}-\frac{ (16\cdot 2^{(4z)}cos(y^{2})ln(2)ln(x))}{9}-\frac{ (20z^{3}cos(y))}{x}}$

${\frac{(20z^{3}cos(y))}{x}-\frac{ (10z^{4}sin(y))}{x}-\frac{ (8\cdot 2^{(4z)}ycos(y^{2})ln(x))}{9}-\frac{ (8\cdot 2^{(4z)}sin(y^{2})ln(2)ln(x))}{9}}$

${(\frac{(20z^{3}cos(y))}{x}-\frac{ (8\cdot 2^{(4z)}sin(y^{2})ln(2)ln(x))}{9})\cdot (\frac{(5z^{4}sin(y))}{x}+\frac{ (4\cdot 2^{(4z)}ycos(y^{2})ln(x))}{9})^{2}}$

Correct answer:

${\frac{(32\cdot 2^{(4z)}y^{2}sin(y^{2})ln(2)ln(x))}{9}-\frac{ (16\cdot 2^{(4z)}cos(y^{2})ln(2)ln(x))}{9}-\frac{ (20z^{3}cos(y))}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(5z^{4}cos(y))}{x}-\frac{ (2\cdot 2^{(4z)}sin(y^{2})ln(x))}{9}\\&f_{y}=-\frac{ (5z^{4}sin(y))}{x}-\frac{ (4\cdot 2^{(4z)}ycos(y^{2})ln(x))}{9}\\&f_{yz}=-\frac{ (20z^{3}sin(y))}{x}-\frac{ (16\cdot 2^{(4z)}ycos(y^{2})ln(2)ln(x))}{9}\\&f_{yzy}=\frac{(32\cdot 2^{(4z)}y^{2}sin(y^{2})ln(2)ln(x))}{9}-\frac{ (16\cdot 2^{(4z)}cos(y^{2})ln(2)ln(x))}{9}-\frac{ (20z^{3}cos(y))}{x}\end{align*}$

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Example Question #789 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xz}\\&\text{Where }f(x,y,z)=3y^{4}cos(z^{2}) -\frac{ (tan(y^{3})ln(x))}{(5z)}\end{align*}$

Possible Answers:

${-\frac{(tan(y^{3})\cdot (\frac{(tan(y^{3})ln(x))}{(5z^{2})}- 6y^{4}zsin(z^{2})))}{(5xz)}}$

${-tan(\frac{y^{3})^{2}}{(25x^{2}z^{3})}}$

${\frac{tan(y^{3})}{(5xz^{2})}}$

${\frac{(tan(y^{3})ln(x))}{(5z^{2})}-\frac{ tan(y^{3})}{(5xz)}- 6y^{4}zsin(z^{2})}$

Correct answer:

${\frac{tan(y^{3})}{(5xz^{2})}}$

Explanation:

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Example Question #790 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\\&\text{Where }f(x,y,z)=\frac{(2e^{(y^{2})})}{(xz^{2})}- sin(y^{2})e^{(z)}cos(4x + x^{2})\end{align*}$

Possible Answers:

${sin(y^{2})e^{(z)}sin(4x + x^{2})\cdot (2x + 4) -\frac{ (2e^{(y^{2})})}{(x^{2}z^{2})}- 2ycos(y^{2})e^{(z)}cos(4x + x^{2}) +\frac{ (4ye^{(y^{2})})}{(xz^{2})}}$

${2ycos(y^{2})e^{(z)}sin(4x + x^{2})\cdot (2x + 4) -\frac{ (4ye^{(y^{2})})}{(x^{2}z^{2})}}$

${(\frac{(4ye^{(y^{2})})}{(x^{2}z^{2})}- 2ycos(y^{2})e^{(z)}sin(4x + x^{2})\cdot (2x + 4))\cdot (\frac{(2e^{(y^{2})})}{(x^{2}z^{2})}- sin(y^{2})e^{(z)}sin(4x + x^{2})\cdot (2x + 4))}$

${(2ycos(y^{2})e^{(z)}cos(4x + x^{2}) -\frac{ (4ye^{(y^{2})})}{(xz^{2})})\cdot (\frac{(2e^{(y^{2})})}{(x^{2}z^{2})}- sin(y^{2})e^{(z)}sin(4x + x^{2})\cdot (2x + 4))}$

Correct answer:

${2ycos(y^{2})e^{(z)}sin(4x + x^{2})\cdot (2x + 4) -\frac{ (4ye^{(y^{2})})}{(x^{2}z^{2})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2e^{(y^{2})})}{(xz^{2})}- sin(y^{2})e^{(z)}cos(4x + x^{2})\\&f_{x}=sin(y^{2})e^{(z)}sin(4x + x^{2})\cdot (2x + 4) -\frac{ (2e^{(y^{2})})}{(x^{2}z^{2})}\\&f_{xy}=2ycos(y^{2})e^{(z)}sin(4x + x^{2})\cdot (2x + 4) -\frac{ (4ye^{(y^{2})})}{(x^{2}z^{2})}\end{align*}$

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Example Question #791 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=\frac{(2ln(3y)e^{(3x)}sin(3z + z^{2}))}{7}\end{align*}$

Possible Answers:

${\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2}))}{7}-\frac{ (2ln(3y)e^{(3x)}sin(3z + z^{2})\cdot (2z + 3)^{2})}{7}}$

${\frac{(2ln(3y)e^{(3x)}cos(3z + z^{2})\cdot (2z + 3)\cdot (\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2}))}{7}-\frac{ (2ln(3y)e^{(3x)}sin(3z + z^{2})\cdot (2z + 3)^{2})}{7}))}{7}}$

${\frac{(4ln(3y)^{2}e^{(6x)}cos(3z + z^{2})^{2}\cdot (2z + 3)^{2})}{49}}$

${\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2})\cdot (2z + 3))}{7}}$

Correct answer:

${\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2}))}{7}-\frac{ (2ln(3y)e^{(3x)}sin(3z + z^{2})\cdot (2z + 3)^{2})}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2ln(3y)e^{(3x)}sin(3z + z^{2}))}{7}\\&f_{z}=\frac{(2ln(3y)e^{(3x)}cos(3z + z^{2})\cdot (2z + 3))}{7}\\&f_{zz}=\frac{(4ln(3y)e^{(3x)}cos(3z + z^{2}))}{7}-\frac{ (2ln(3y)e^{(3x)}sin(3z + z^{2})\cdot (2z + 3)^{2})}{7}\end{align*}$

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Example Question #792 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zzy}\\&\text{Where }f(x,y,z)=5y^{2}tan(x^{2})e^{(z)} - 2^{(3z)}e^{(4x)}e^{(4y)}\end{align*}$

Possible Answers:

${(5y^{2}tan(x^{2})e^{(z)} - 3\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2))\cdot (10ytan(x^{2})e^{(z)} - 36\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2})\cdot (5y^{2}tan(x^{2})e^{(z)} - 9\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2})}$

${10ytan(x^{2})e^{(z)} - 4\cdot 2^{(3z)}e^{(4x)}e^{(4y)} + 10y^{2}tan(x^{2})e^{(z)} - 6\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)}$

${(5y^{2}tan(x^{2})e^{(z)} - 3\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2))^{2}\cdot (10ytan(x^{2})e^{(z)} - 4\cdot 2^{(3z)}e^{(4x)}e^{(4y)})}$

${10ytan(x^{2})e^{(z)} - 36\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2}}$

Correct answer:

${10ytan(x^{2})e^{(z)} - 36\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=5y^{2}tan(x^{2})e^{(z)} - 2^{(3z)}e^{(4x)}e^{(4y)}\\&f_{z}=5y^{2}tan(x^{2})e^{(z)} - 3\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)\\&f_{zz}=5y^{2}tan(x^{2})e^{(z)} - 9\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2}\\&f_{zzy}=10ytan(x^{2})e^{(z)} - 36\cdot 2^{(3z)}e^{(4x)}e^{(4y)}ln(2)^{2}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #783 : Partial Derivatives

Example Question #784 : Partial Derivatives

Example Question #785 : Partial Derivatives

Example Question #786 : Partial Derivatives

Example Question #787 : Partial Derivatives

Example Question #788 : Partial Derivatives

Example Question #789 : Partial Derivatives

Example Question #790 : Partial Derivatives

Example Question #791 : Partial Derivatives

Example Question #792 : Partial Derivatives

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