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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #631 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #632 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=\frac{(2\cdot2^{(3y)}cos(4x + x^{2})ln(z))}{7}- 4ln(y^{2})sin(z^{3})\end{align*}$

Possible Answers:

${36z^{4}ln(y^{2})sin(z^{3}) - 24zcos(z^{3})ln(y^{2}) -\frac{ (2\cdot2^{(3y)}cos(4x + x^{2}))}{(7z^{2})}}$

${\frac{(4\cdot2^{(3y)}cos(4x + x^{2}))}{(7z)}- 24z^{2}cos(z^{3})ln(y^{2})}$

${(12z^{2}cos(z^{3})ln(y^{2}) -\frac{ (2\cdot2^{(3y)}cos(4x + x^{2}))}{(7z)})^{2}}$

${(12z^{2}cos(z^{3})ln(y^{2}) -\frac{ (2\cdot2^{(3y)}cos(4x + x^{2}))}{(7z)})\cdot(24zcos(z^{3})ln(y^{2}) - 36z^{4}ln(y^{2})sin(z^{3}) +\frac{ (2\cdot2^{(3y)}cos(4x + x^{2}))}{(7z^{2})})}$

Correct answer:

${36z^{4}ln(y^{2})sin(z^{3}) - 24zcos(z^{3})ln(y^{2}) -\frac{ (2\cdot2^{(3y)}cos(4x + x^{2}))}{(7z^{2})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot2^{(3y)}cos(4x + x^{2})ln(z))}{7}- 4ln(y^{2})sin(z^{3})\\&f_{z}=\frac{(2\cdot2^{(3y)}cos(4x + x^{2}))}{(7z)}- 12z^{2}cos(z^{3})ln(y^{2})\\&f_{zz}=36z^{4}ln(y^{2})sin(z^{3}) - 24zcos(z^{3})ln(y^{2}) -\frac{ (2\cdot2^{(3y)}cos(4x + x^{2}))}{(7z^{2})}\end{align*}$

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Example Question #633 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{(ln(4z)cos(x)ln(y))}{4}\end{align*}$

Possible Answers:

${-\frac{(ln(4z)^{2}cos(x)sin(x))}{(16y^{2})}}$

${-\frac{(ln(4z)sin(x))}{(4y)}}$

${-\frac{(ln(4z)^{2}cos(x)ln(y)sin(x))}{(16y)}}$

${\frac{(ln(4z)cos(x))}{(4y)}-\frac{ (ln(4z)ln(y)sin(x))}{4}}$

Correct answer:

${-\frac{(ln(4z)sin(x))}{(4y)}}$

Explanation:

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

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Example Question #634 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #635 : Partial Derivatives

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

Possible Answers:

${\frac{(16e^{(8x)}e^{(8z)})}{(25y^{2})}}$

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

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

${\frac{(8e^{(4x)}e^{(4z)})}{(5y)}}$

Correct answer:

${\frac{(8e^{(4x)}e^{(4z)})}{(5y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(ln(y^{2})e^{(4x)}e^{(4z)})}{5}\\&f_{y}=\frac{(2e^{(4x)}e^{(4z)})}{(5y)}\\&f_{yz}=\frac{(8e^{(4x)}e^{(4z)})}{(5y)}\end{align*}$

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Example Question #634 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zy}\\&\text{Where }f(x,y,z)=\frac{(4y^{4}cos(z^{2}))}{x^{2}}+ 3\cdot2^z\cdot4^{(4x)}e^{(3y)}\end{align*}$

Possible Answers:

${(3\cdot2^z\cdot4^{(4x)}e^{(3y)}ln(2) -\frac{ (8y^{4}zsin(z^{2}))}{x^{2}})\cdot(\frac{(16y^{3}cos(z^{2}))}{x^{2}}+ 9\cdot2^z\cdot4^{(4x)}e^{(3y)})}$

${\frac{(16y^{3}cos(z^{2}))}{x^{2}}+ 9\cdot2^z\cdot4^{(4x)}e^{(3y)} + 3\cdot2^z\cdot4^{(4x)}e^{(3y)}ln(2) -\frac{ (8y^{4}zsin(z^{2}))}{x^{2}}}$

${9\cdot2^z\cdot4^{(4x)}e^{(3y)}ln(2) -\frac{ (32y^{3}zsin(z^{2}))}{x^{2}}}$

${(3\cdot2^z\cdot4^{(4x)}e^{(3y)}ln(2) -\frac{ (8y^{4}zsin(z^{2}))}{x^{2}})\cdot(9\cdot2^z\cdot4^{(4x)}e^{(3y)}ln(2) -\frac{ (32y^{3}zsin(z^{2}))}{x^{2}})}$

Correct answer:

${9\cdot2^z\cdot4^{(4x)}e^{(3y)}ln(2) -\frac{ (32y^{3}zsin(z^{2}))}{x^{2}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(4y^{4}cos(z^{2}))}{x^{2}}+ 3\cdot2^z\cdot4^{(4x)}e^{(3y)}\\&f_{z}=3\cdot2^z\cdot4^{(4x)}e^{(3y)}ln(2) -\frac{ (8y^{4}zsin(z^{2}))}{x^{2}}\\&f_{zy}=9\cdot2^z\cdot4^{(4x)}e^{(3y)}ln(2) -\frac{ (32y^{3}zsin(z^{2}))}{x^{2}}\end{align*}$

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Example Question #635 : Partial Derivatives

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

Possible Answers:

${-(45y^{2}z^{2}cos(y^{3})cos(z^{3})e^{(x)} -\frac{ (2\cdot4^{(y^{2})}yln(4)cos(4x + x^{3}))}{(5z)})\cdot(\frac{(4^{(y^{2})}cos(4x + x^{3}))}{(5z)}- 15z^{2}cos(z^{3})sin(y^{3})e^{(x)})}$

${\frac{(4^{(y^{2})}cos(4x + x^{3}))}{(5z)}- 15y^{2}cos(y^{3})sin(z^{3})e^{(x)} - 15z^{2}cos(z^{3})sin(y^{3})e^{(x)} +\frac{ (2\cdot4^{(y^{2})}yln(4z)ln(4)cos(4x + x^{3}))}{5}}$

${-(15y^{2}cos(y^{3})sin(z^{3})e^{(x)} -\frac{ (2\cdot4^{(y^{2})}yln(4z)ln(4)cos(4x + x^{3}))}{5})\cdot(\frac{(4^{(y^{2})}cos(4x + x^{3}))}{(5z)}- 15z^{2}cos(z^{3})sin(y^{3})e^{(x)})}$

${\frac{(2\cdot4^{(y^{2})}yln(4)cos(4x + x^{3}))}{(5z)}- 45y^{2}z^{2}cos(y^{3})cos(z^{3})e^{(x)}}$

Correct answer:

${\frac{(2\cdot4^{(y^{2})}yln(4)cos(4x + x^{3}))}{(5z)}- 45y^{2}z^{2}cos(y^{3})cos(z^{3})e^{(x)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(4^{(y^{2})}ln(4z)cos(4x + x^{3}))}{5}- 5sin(y^{3})sin(z^{3})e^{(x)}\\&f_{z}=\frac{(4^{(y^{2})}cos(4x + x^{3}))}{(5z)}- 15z^{2}cos(z^{3})sin(y^{3})e^{(x)}\\&f_{zy}=\frac{(2\cdot4^{(y^{2})}yln(4)cos(4x + x^{3}))}{(5z)}- 45y^{2}z^{2}cos(y^{3})cos(z^{3})e^{(x)}\end{align*}$

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Example Question #636 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zz}\\&\text{Where }f(x,y,z)=-\frac{ (2\cdot3^{(3z)}e^{(4y)}sin(x))}{9}- x^{4}ycos(z^{2})\end{align*}$

Possible Answers:

${4x^{4}yzsin(z^{2}) -\frac{ (4\cdot3^{(3z)}e^{(4y)}ln(3)sin(x))}{3}}$

${-(\frac{(2\cdot3^{(3z)}e^{(4y)}ln(3)sin(x))}{3}- 2x^{4}yzsin(z^{2}))\cdot(2x^{4}ysin(z^{2}) - 2\cdot3^{(3z)}e^{(4y)}ln(3)^{2}sin(x) + 4x^{4}yz^{2}cos(z^{2}))}$

${(\frac{(2\cdot3^{(3z)}e^{(4y)}ln(3)sin(x))}{3}- 2x^{4}yzsin(z^{2}))^{2}}$

${2x^{4}ysin(z^{2}) - 2\cdot3^{(3z)}e^{(4y)}ln(3)^{2}sin(x) + 4x^{4}yz^{2}cos(z^{2})}$

Correct answer:

${2x^{4}ysin(z^{2}) - 2\cdot3^{(3z)}e^{(4y)}ln(3)^{2}sin(x) + 4x^{4}yz^{2}cos(z^{2})}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{ (2\cdot3^{(3z)}e^{(4y)}sin(x))}{9}- x^{4}ycos(z^{2})\\&f_{z}=2x^{4}yzsin(z^{2}) -\frac{ (2\cdot3^{(3z)}e^{(4y)}ln(3)sin(x))}{3}\\&f_{zz}=2x^{4}ysin(z^{2}) - 2\cdot3^{(3z)}e^{(4y)}ln(3)^{2}sin(x) + 4x^{4}yz^{2}cos(z^{2})\end{align*}$

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Example Question #636 : Partial Derivatives

Find the partial derivative of xy^2+xz+e^y with the respect to .

Possible Answers:

y^2+z

3x+1

y^2+z+e^y

2xy+z+1

2x+z+1

Correct answer:

y^2+z

Explanation:

Assume that all other variables besides are constants.

Take the derivative of each term accordingly.

$\frac{\partial }{\partial x} (xy^2) = y^2$

$\frac{\partial }{\partial x} (xz) = z$

$\frac{\partial }{\partial x} (e^y) =0$

Sum all the terms.

The answer is: y^2+z

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Example Question #637 : Partial Derivatives

Find the partial derivative f_x of $\small f(x,y)=2xy+5\pi x^{3/2}$ .

Possible Answers:

$\small \small f_x=2x+\frac{15\pi}{2} \sqrt{x^3}$

$\small f_x=2y+\frac{15\pi}{2} \sqrt{x}$

$\small \small f_x=2x+\frac{15\pi}{2} \sqrt{x}$

$\small \small f_x=2xy+\frac{15\pi}{2} \sqrt{x^3}$

Correct answer:

$\small f_x=2y+\frac{15\pi}{2} \sqrt{x}$

Explanation:

To find the partial derivative f_x of $\small f(x,y)=2xy+5\pi x^{3/2}$ , we need to take its derivative with respect to $\small x$ , which means holding $\small y$ constant. So we use the power rule

$\small (x^a)'=ax^{a-1}$

to get

$\small \small f_x =\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}[2xy+5\pi x^{3/2}]=\frac{\partial}{\partial x}2xy+\frac{\partial}{\partial x}5\pi x^{3/2}$

$\small =2y+5\cdot (3/2)\pi x^{1/2}=2y+\frac{15\pi}{2}\sqrt{x}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #631 : Partial Derivatives

Example Question #632 : Partial Derivatives

Example Question #633 : Partial Derivatives

Example Question #634 : Partial Derivatives

Example Question #635 : Partial Derivatives

Example Question #634 : Partial Derivatives

Example Question #635 : Partial Derivatives

Example Question #636 : Partial Derivatives

Example Question #636 : Partial Derivatives

Example Question #637 : Partial Derivatives

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