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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #2859 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xzy}\\&\text{Where }f(x,y,z)=-10sin{(y^4)}e^{(z^4)}e^{(x)}\end{align*}$

Possible Answers:

${- 10sin{(y^4)}e^{(z^4)}e^{(x)} - 40y^3cos{(y^4)}e^{(z^4)}e^{(x)} - 40z^3sin{(y^4)}e^{(z^4)}e^{(x)}}$

${-160y^3z^3cos{(y^4)}e^{(z^4)}e^{(x)}}$

$-16000y^3z^3cos{(y^4)}sin{(y^4)}^2e^{(3x)}e^{(3z^4)}$

${- 10sin{(y^4)}e^{(z^4)}e^{(x)} - 40z^3sin{(y^4)}e^{(z^4)}e^{(x)} - 160y^3z^3cos{(y^4)}e^{(z^4)}e^{(x)}}$

Correct answer:

${-160y^3z^3cos{(y^4)}e^{(z^4)}e^{(x)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f_{x}=-10sin{(y^4)}e^{(z^4)}e^{(x)}\\&f_{xz}=-40z^3sin{(y^4)}e^{(z^4)}e^{(x)}\\&f_{xzy}=-160y^3z^3cos{(y^4)}e^{(z^4)}e^{(x)}\end{align*}$

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Example Question #2860 : Calculus 3

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

Possible Answers:

${\frac{{(12x^3y^2sin{(x^4)})}}{z^2}}$

${\frac{-{(48x^6y^8cos{(x^4)}sin{(x^4)}^2)}}{z^6}}$

$\frac{{(12x^3y^8cos{(x^4)}^2sin{(x^4)})}}{z^4}$

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

Correct answer:

${\frac{{(12x^3y^2sin{(x^4)})}}{z^2}}$

Explanation:

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

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Example Question #2861 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yxy}\\&\text{Where }f(x,y,z)=\frac{-{(7\cdot3^ycos{(x^2)})}}{z}\end{align*}$

Possible Answers:

$\frac{{(686\cdot3^{(3y)}xcos{(x^2)}^2sin{(x^2)}ln{(3)}^2)}}{z^3}$

${\frac{{(14\cdot3^yxsin{(x^2)}ln{(3)}^2)}}{z}}$

${\frac{{(14\cdot3^yxsin{(x^2)})}}{z} -\frac{ {(14\cdot3^ycos{(x^2)}ln{(3)})}}{z}}$

${\frac{-{(1372\cdot3^{(3y)}x^2cos{(x^2)}sin{(x^2)}^2ln{(3)}^4)}}{z^3}}$

Correct answer:

${\frac{{(14\cdot3^yxsin{(x^2)}ln{(3)}^2)}}{z}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f_{y}=\frac{-{(7\cdot3^ycos{(x^2)}ln{(3)})}}{z}\\&f_{yx}=\frac{{(14\cdot3^yxsin{(x^2)}ln{(3)})}}{z}\\&f_{yxy}=\frac{{(14\cdot3^yxsin{(x^2)}ln{(3)}^2)}}{z}\end{align*}$

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Example Question #2862 : Calculus 3

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

Possible Answers:

${8yz^2e^{(x^3)} + 8y^2ze^{(x^3)}}$

$64y^3z^3e^{(2x^3)}$

${16yze^{(x^3)}}$

${128y^3z^2e^{(2x^3)}}$

Correct answer:

${16yze^{(x^3)}}$

Explanation:

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Example Question #2863 : Calculus 3

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

Possible Answers:

$\frac{{(18\cdot2^{(2x^4)}ze^{(2z^2)}ln{(y)})}}{y}$

${\frac{{(18\cdot2^{(2x^4)}ze^{(2z^2)})}}{y^2}}$

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

${\frac{{(6\cdot2^{(x^4)}ze^{(z^2)})}}{y}}$

Correct answer:

${\frac{{(6\cdot2^{(x^4)}ze^{(z^2)})}}{y}}$

Explanation:

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

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Example Question #2864 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yzy}\\&\text{Where }f(x,y,z)=-8\cdot4^y\cdot4^zsin{(x^4)}\end{align*}$

Possible Answers:

${-512\cdot4^{(3y)}\cdot4^{(3z)}sin{(x^4)}^3ln{(4)}^6}$

${-24\cdot4^y\cdot4^zsin{(x^4)}ln{(4)}}$

${-8\cdot4^y\cdot4^zsin{(x^4)}ln{(4)}^3}$

$-512\cdot4^{(3y)}\cdot4^{(3z)}sin{(x^4)}^3ln{(4)}^3$

Correct answer:

${-8\cdot4^y\cdot4^zsin{(x^4)}ln{(4)}^3}$

Explanation:

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

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Example Question #2865 : Calculus 3

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

Possible Answers:

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

${\frac{-{(16e^{(4x)}ln{(y)}^2)}}{{(y^3z^8)}}}$

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

$\frac{{(16e^{(4x)}ln{(y)}^2)}}{{(y^2z^8)}}$

Correct answer:

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

Explanation:

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

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Example Question #2866 : Calculus 3

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

Possible Answers:

${\frac{-{(80x^3cos{(z^2)}sin{(x^4)})}}{y}}$

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

$\frac{-{(400x^3cos{(x^4)}cos{(z^2)}^2ln{(y^4)}sin{(x^4)})}}{y}$

${\frac{-{(1600x^3cos{(x^4)}cos{(z^2)}^2sin{(x^4)})}}{y^2}}$

Correct answer:

${\frac{-{(80x^3cos{(z^2)}sin{(x^4)})}}{y}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f_{y}=\frac{{(20cos{(x^4)}cos{(z^2)})}}{y}\\&f_{yx}=\frac{-{(80x^3cos{(z^2)}sin{(x^4)})}}{y}\end{align*}$

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Example Question #2867 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=-10sin{(y^4)}ln{(x)}sin{(z)}\end{align*}$

Possible Answers:

${\frac{-{(40y^3cos{(y^4)}sin{(z)})}}{x}}$

${-\frac{ {(10sin{(y^4)}sin{(z)})}}{x} - 40y^3cos{(y^4)}ln{(x)}sin{(z)}}$

${\frac{{(1600y^6cos{(y^4)}^2ln{(x)}sin{(z)}^2)}}{x}}$

$\frac{{(400y^3cos{(y^4)}sin{(y^4)}ln{(x)}sin{(z)}^2)}}{x}$

Correct answer:

${\frac{-{(40y^3cos{(y^4)}sin{(z)})}}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f_{y}=-40y^3cos{(y^4)}ln{(x)}sin{(z)}\\&f_{yx}=\frac{-{(40y^3cos{(y^4)}sin{(z)})}}{x}\end{align*}$

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Example Question #2868 : Calculus 3

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

Possible Answers:

${\frac{{(256sin{(x^3)}^4e^{(4y)}ln{(z)})}}{z^5}}$

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

$\frac{{(256sin{(x^3)}^4e^{(4y)}ln{(z)}^2)}}{z^2}$

${\frac{{(8sin{(x^3)}e^{(y)})}}{z} + 8sin{(x^3)}e^{(y)}ln{(z)}}$

Correct answer:

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

Explanation:

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

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #2859 : Calculus 3

Example Question #2860 : Calculus 3

Example Question #2861 : Calculus 3

Example Question #2862 : Calculus 3

Example Question #2863 : Calculus 3

Example Question #2864 : Calculus 3

Example Question #2865 : Calculus 3

Example Question #2866 : Calculus 3

Example Question #2867 : Calculus 3

Example Question #2868 : Calculus 3

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