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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #581 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xxzy}\\&\text{Where }f(x,y,z)=\frac{{(2\cdot4^{(3y)}x^2cos{(z)})}}{7}\end{align*}$

Possible Answers:

${\frac{{(8\cdot4^{(3y)}xcos{(z)})}}{7} -\frac{ {(2\cdot4^{(3y)}x^2sin{(z)})}}{7} +\frac{ {(6\cdot4^{(3y)}x^2ln{(4)}cos{(z)})}}{7}}$

${\frac{-{(192\cdot4^{(12y)}x^6ln{(4)}cos{(z)}^3sin{(z)})}}{2401}}$

${\frac{{(768\cdot4^{(12y)}xln{(4)}cos{(z)}^2sin{(z)}^2)}}{2401}}$

${\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}}$

Correct answer:

${\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2\cdot4^{(3y)}x^2cos{(z)})}}{7}\\&f_{x}=\frac{{(4\cdot4^{(3y)}xcos{(z)})}}{7}\\&f_{xx}=\frac{{(4\cdot4^{(3y)}cos{(z)})}}{7}\\&f_{xxz}=\frac{-{(4\cdot4^{(3y)}sin{(z)})}}{7}\\&f_{xxzy}=\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}\end{align*}$

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Example Question #582 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xx}\\&\text{Where }f(x,y,z)=2\cdot3^{(3x)}\cdot3^ysin{(z^2)}\end{align*}$

Possible Answers:

${36\cdot3^{(6x)}\cdot3^{(2y)}sin{(z^2)}^2ln{(3)}^2}$

${108\cdot3^{(6x)}\cdot3^{(2y)}sin{(z^2)}^2ln{(3)}^3}$

${12\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}}$

${18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2}$

Correct answer:

${18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=2\cdot3^{(3x)}\cdot3^ysin{(z^2)}\\&f_{x}=6\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}\\&f_{xx}=18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2\end{align*}$

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Example Question #583 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xx}\\&\text{Where }f(x,y,z)=2\cdot3^{(3x)}\cdot3^ysin{(z^2)}\end{align*}$

Possible Answers:

${36\cdot3^{(6x)}\cdot3^{(2y)}sin{(z^2)}^2ln{(3)}^2}$

${12\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}}$

${18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2}$

${108\cdot3^{(6x)}\cdot3^{(2y)}sin{(z^2)}^2ln{(3)}^3}$

Correct answer:

${18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=2\cdot3^{(3x)}\cdot3^ysin{(z^2)}\\&f_{x}=6\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}\\&f_{xx}=18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2\end{align*}$

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Example Question #584 : Partial Derivatives

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

Possible Answers:

${\frac{-{(3\cdot3^{(2y^2)}sin{(4z)}^2)}}{{(4x^7)}}}$

${\frac{{(3^{(2y^2)}sin{(4z)}^2)}}{{(4x^6)}}}$

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

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

Correct answer:

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

Explanation:

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

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Example Question #585 : Partial Derivatives

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

Possible Answers:

{-6xcos{(3y)}ln{(4z)}}

${-6xcos{(3y)}ln{(4z)}\cdot{(4y^3e^{(z^2)} - 3x^2cos{(3y)}ln{(4z)})}}$

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

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

Correct answer:

{-6xcos{(3y)}ln{(4z)}}

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=y^4e^{(z^2)} - x^2ln{(4z)}sin{(3y)}\\&f_{y}=4y^3e^{(z^2)} - 3x^2cos{(3y)}ln{(4z)}\\&f_{yx}=-6xcos{(3y)}ln{(4z)}\end{align*}$

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Example Question #2951 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yy}\\&\text{Where }f(x,y,z)=3z^3sin{(y^3)}ln{(x)} -\frac{ {(2\cdot2^xy^2cos{(4z)})}}{9}\end{align*}$

Possible Answers:

${18y^2z^3cos{(y^3)}ln{(x)} -\frac{ {(8\cdot2^xycos{(4z)})}}{9}}$

${18yz^3cos{(y^3)}ln{(x)} - 27y^4z^3sin{(y^3)}ln{(x)} -\frac{ {(4\cdot2^xcos{(4z)})}}{9}}$

${18yz^3cos{(y^3)}ln{(x)} + 27y^4z^3sin{(y^3)}ln{(x)} +\frac{ {(4\cdot2^xcos{(4z)})}}{9}}$

${18y^2z^3cos{(y^3)}ln{(x)} +\frac{ {(8\cdot2^xycos{(4z)})}}{9}}$

Correct answer:

${18yz^3cos{(y^3)}ln{(x)} - 27y^4z^3sin{(y^3)}ln{(x)} -\frac{ {(4\cdot2^xcos{(4z)})}}{9}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3z^3sin{(y^3)}ln{(x)} -\frac{ {(2\cdot2^xy^2cos{(4z)})}}{9}\\&f_{y}=9y^2z^3cos{(y^3)}ln{(x)} -\frac{ {(4\cdot2^xycos{(4z)})}}{9}\\&f_{yy}=18yz^3cos{(y^3)}ln{(x)} - 27y^4z^3sin{(y^3)}ln{(x)} -\frac{ {(4\cdot2^xcos{(4z)})}}{9}\end{align*}$

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Example Question #587 : Partial Derivatives

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

Possible Answers:

${{15y^2e^{(z^2)} -\frac{ {(2ysin{(3z)}e^{(3x)}e^{(y^2)})}}{5}^2}}$

${30y^2e^{(z^2)} -\frac{ {(4ysin{(3z)}e^{(3x)}e^{(y^2)})}}{5}}$

${30ye^{(z^2)} -\frac{ {(2sin{(3z)}e^{(3x)}e^{(y^2)})}}{5} -\frac{ {(4y^2sin{(3z)}e^{(3x)}e^{(y^2)})}}{5}}$

${30y^2e^{(z^2)}+\frac{ {(4ysin{(3z)}e^{(3x)}e^{(y^2)})}}{5}}$

Correct answer:

${30ye^{(z^2)} -\frac{ {(2sin{(3z)}e^{(3x)}e^{(y^2)})}}{5} -\frac{ {(4y^2sin{(3z)}e^{(3x)}e^{(y^2)})}}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=5y^3e^{(z^2)} -\frac{ {(sin{(3z)}e^{(3x)}e^{(y^2)})}}{5}\\&f_{y}=15y^2e^{(z^2)} -\frac{ {(2ysin{(3z)}e^{(3x)}e^{(y^2)})}}{5}\\&f_{yy}=30ye^{(z^2)} -\frac{ {(2sin{(3z)}e^{(3x)}e^{(y^2)})}}{5} -\frac{ {(4y^2sin{(3z)}e^{(3x)}e^{(y^2)})}}{5}\end{align*}$

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Example Question #583 : Partial Derivatives

Find f_y (x,y) given $f(x,y)= x^{y+x}$

Possible Answers:

$\ln x$

$x^{x+1}$

$\ln x \cdot x^{y+x}$

$\ln y \cdot x^{y+x}$

Correct answer:

$\ln x \cdot x^{y+x}$

Explanation:

This derivative is most easily done using logarithmic differentiation. Consider the following

$\begin{align*} f(x,y)&=x^{y+x} \\ \ln f&= \ln x \cdot (y+x) \\ \ln f&= \ln x \cdot y + \ln x \cdot x \\ \frac{f_y}{f}&= \ln x\\ f_y &= f \cdot \ln x = \ln x \cdot x^{y+x} \end{align*}$

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Example Question #2952 : Calculus 3

Find $\frac{\partial f}{\partial z}$ given $f(x,y,z)= -\frac{x^2+y^2}{\cos z}$

Possible Answers:

$\left( x^2+y^2\right )\tan z$

$\frac{\tan z}{\cos z}$

$\frac{\left( x^2+y^2\right )\tan z}{\cos z}$

$\frac{\left( x^2+y^2\right )\sin z}{\cos z}$

Correct answer:

$\frac{\left( x^2+y^2\right )\tan z}{\cos z}$

Explanation:

While taking a derivative with respect to only one variable, the other variables are treated as constants. Consider the following

$\begin{align*} \frac{\partial f}{\partial z}&= \frac{\partial}{\partial z} \left( -\frac{x^2+y^2}{\cos z}\right )\\ &= -\left(x^2+y^2 \right )\frac{\partial }{\partial z} \left(\cos z \right )^{-1}\\ &=- \left(x^2+y^2 \right )\frac{-\sin z}{\cos ^2 z} \\ \frac{\partial f}{\partial z}&= \frac{\left(x^2+y^2 \right ) \tan z}{\cos z} \end{align*}$

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Example Question #2951 : Calculus 3

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

Possible Answers:

${\frac{{(2\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }+\frac{ {(4\cdot2^{{({4x)}}}z^{2}e^{{({4y)}}})}}{5 }-\frac{ {(3e^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2} }+\frac{ {(6ye^{{({x^{2})}}}e^{{({y^{2})}}})}}{z}}$

${\frac{{(2\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }-\frac{ {(4\cdot2^{{({4x)}}}z^{2}e^{{({4y)}}})}}{5 }-\frac{ {(3e^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2} }+\frac{ {(6ye^{{({x^{2})}}}e^{{({y^{2})}}})}}{z}}$

${\frac{{(8\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }-\frac{ {(6ye^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2}}}$

${\frac{{(8\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }+\frac{ {(6ye^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2}}}$

Correct answer:

${\frac{{(8\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }-\frac{ {(6ye^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2^{{({4x)}}}z^{2}e^{{({4y)}}})}}{5 }+\frac{ {(3e^{{({x^{2})}}}e^{{({y^{2})}}})}}{z}\\&f_{z}=\frac{{(2\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }-\frac{ {(3e^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2}}\\&f_{zy}=\frac{{(8\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }-\frac{ {(6ye^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2}}\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #581 : Partial Derivatives

Example Question #582 : Partial Derivatives

Example Question #583 : Partial Derivatives

Example Question #584 : Partial Derivatives

Example Question #585 : Partial Derivatives

Example Question #2951 : Calculus 3

Example Question #587 : Partial Derivatives

Example Question #583 : Partial Derivatives

Example Question #2952 : Calculus 3

Example Question #2951 : Calculus 3

All Calculus 3 Resources

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