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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #2889 : Calculus 3

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

Possible Answers:

${196608cos{(4x)}^2cos{(3y)}sin{(4x)}^2sin{(3y)}^3cos{(z)}^3sin{(z)}}$

{- 16cos{(4x)}sin{(3y)}sin{(z)} - 6cos{(3y)}sin{(4x)}sin{(z)} - 2sin{(4x)}sin{(3y)}cos{(z)}}

${768cos{(4x)}^2cos{(3y)}sin{(4x)}^2sin{(3y)}^3cos{(z)}sin{(z)}^3}$

{96cos{(3y)}sin{(4x)}cos{(z)}}

Correct answer:

{96cos{(3y)}sin{(4x)}cos{(z)}}

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f_{x}=-8cos{(4x)}sin{(3y)}sin{(z)}\\&f_{xz}=-8cos{(4x)}sin{(3y)}cos{(z)}\\&f_{xzx}=32sin{(4x)}sin{(3y)}cos{(z)}\\&f_{xzxy}=96cos{(3y)}sin{(4x)}cos{(z)}\end{align*}$

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Example Question #2890 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\\&\text{Where }f(x,y,z)=7z^2e^{(3x)}e^{(4y)}\end{align*}$

Possible Answers:

${84z^2e^{(3x)}e^{(4y)}}$

${588z^4e^{(6x)}e^{(8y)}}$

${1764z^4e^{(6x)}e^{(8y)}}$

${49z^2e^{(3x)}e^{(4y)}}$

Correct answer:

${84z^2e^{(3x)}e^{(4y)}}$

Explanation:

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Example Question #2891 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{{(6ln{(4y)}e^{(3x)})}}{z^2}\end{align*}$

Possible Answers:

${\frac{{(108e^{(6x)})}}{{(y^2z^4)}}}$

${\frac{{(6e^{(3x)})}}{{(yz^2)}} +\frac{ {(18ln{(4y)}e^{(3x)})}}{z^2}}$

${\frac{{(108ln{(4y)}e^{(6x)})}}{{(yz^4)}}}$

${\frac{{(18e^{(3x)})}}{{(yz^2)}}}$

Correct answer:

${\frac{{(18e^{(3x)})}}{{(yz^2)}}}$

Explanation:

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

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Example Question #2892 : Calculus 3

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

Possible Answers:

${\frac{-{(288\cdot3^{(4x)}zln{(3)}^2)}}{y}}$

${\frac{{(107495424\cdot3^{(16x)}z^4ln{(3y)}^3ln{(3)}^5)}}{y}}$

${-\frac{ {(9\cdot3^{(4x)}z^2)}}{y} - 18\cdot3^{(4x)}zln{(3y)} - 72\cdot3^{(4x)}z^2ln{(3y)}ln{(3)}}$

${\frac{{(209952\cdot3^{(16x)}z^7ln{(3y)}^3ln{(3)}^2)}}{y}}$

Correct answer:

${\frac{-{(288\cdot3^{(4x)}zln{(3)}^2)}}{y}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f_{z}=-18\cdot3^{(4x)}zln{(3y)}\\&f_{zx}=-72\cdot3^{(4x)}zln{(3y)}ln{(3)}\\&f_{zxx}=-288\cdot3^{(4x)}zln{(3y)}ln{(3)}^2\\&f_{zxxy}=\frac{-{(288\cdot3^{(4x)}zln{(3)}^2)}}{y}\end{align*}$

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Example Question #2893 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zyzx}\\&\text{Where }f(x,y,z)=-2\cdot3^{(4x)}\cdot4^ze^{(3y)}\end{align*}$

Possible Answers:

${-24\cdot3^{(4x)}\cdot4^ze^{(3y)}ln{(3)}ln{(4)}^2}$

${192\cdot3^{(16x)}\cdot4^{(4z)}e^{(12y)}ln{(3)}ln{(4)}^2}$

${- 6\cdot3^{(4x)}\cdot4^ze^{(3y)} - 8\cdot3^{(4x)}\cdot4^ze^{(3y)}ln{(3)} - 4\cdot3^{(4x)}\cdot4^ze^{(3y)}ln{(4)}}$

${1728\cdot3^{(16x)}\cdot4^{(4z)}e^{(12y)}ln{(3)}ln{(4)}^6}$

Correct answer:

${-24\cdot3^{(4x)}\cdot4^ze^{(3y)}ln{(3)}ln{(4)}^2}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f_{z}=-2\cdot3^{(4x)}\cdot4^ze^{(3y)}ln{(4)}\\&f_{zy}=-6\cdot3^{(4x)}\cdot4^ze^{(3y)}ln{(4)}\\&f_{zyz}=-6\cdot3^{(4x)}\cdot4^ze^{(3y)}ln{(4)}^2\\&f_{zyzx}=-24\cdot3^{(4x)}\cdot4^ze^{(3y)}ln{(3)}ln{(4)}^2\end{align*}$

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Example Question #2894 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yyzz}\\&\text{Where }f(x,y,z)=-3\cdot3^{(4z)}cos{(4x)}ln{(3y)}\end{align*}$

Possible Answers:

${\frac{-{(5184\cdot3^{(16z)}cos{(4x)}^4ln{(3)}^3)}}{y^7}}$

${\frac{{(1296\cdot3^{(16z)}cos{(4x)}^4ln{(3y)}^2ln{(3)}^2)}}{y^2}}$

${-\frac{ {(6\cdot3^{(4z)}cos{(4x)})}}{y} - 24\cdot3^{(4z)}cos{(4x)}ln{(3y)}ln{(3)}}$

${\frac{{(48\cdot3^{(4z)}cos{(4x)}ln{(3)}^2)}}{y^2}}$

Correct answer:

${\frac{{(48\cdot3^{(4z)}cos{(4x)}ln{(3)}^2)}}{y^2}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f_{y}=\frac{-{(3\cdot3^{(4z)}cos{(4x)})}}{y}\\&f_{yy}=\frac{{(3\cdot3^{(4z)}cos{(4x)})}}{y^2}\\&f_{yyz}=\frac{{(12\cdot3^{(4z)}cos{(4x)}ln{(3)})}}{y^2}\\&f_{yyzz}=\frac{{(48\cdot3^{(4z)}cos{(4x)}ln{(3)}^2)}}{y^2}\end{align*}$

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Example Question #2895 : Calculus 3

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

Possible Answers:

${\frac{{(8\cdot2^{(3z)})}}{{(x^3y^6)}}}$

${\frac{{(6\cdot2^z)}}{{(xy^2)}}}$

${\frac{-{(16\cdot2^{(3z)})}}{{(x^6y^6)}}}$

${\frac{{(4\cdot2^z)}}{{(x^3y^2)}}}$

Correct answer:

${\frac{{(4\cdot2^z)}}{{(x^3y^2)}}}$

Explanation:

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

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Example Question #2896 : Calculus 3

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

Possible Answers:

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

${\frac{{(10y^2e^{(x)})}}{z}}$

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

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

Correct answer:

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

Explanation:

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

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Example Question #2897 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{-{(10\cdot3^{(3z)}\cdot4^{(4y)})}}{x^2}\end{align*}$

Possible Answers:

${\frac{{(20\cdot3^{(3z)}\cdot4^{(4y)})}}{x^3} -\frac{ {(40\cdot3^{(3z)}\cdot4^{(4y)}ln{(4)})}}{x^2}}$

${\frac{-{(800\cdot3^{(6z)}\cdot4^{(8y)}ln{(4)})}}{x^5}}$

${\frac{{(80\cdot3^{(3z)}\cdot4^{(4y)}ln{(4)})}}{x^3}}$

${\frac{-{(3200\cdot3^{(6z)}\cdot4^{(8y)}ln{(4)}^2)}}{x^5}}$

Correct answer:

${\frac{{(80\cdot3^{(3z)}\cdot4^{(4y)}ln{(4)})}}{x^3}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f_{y}=\frac{-{(40\cdot3^{(3z)}\cdot4^{(4y)}ln{(4)})}}{x^2}\\&f_{yx}=\frac{{(80\cdot3^{(3z)}\cdot4^{(4y)}ln{(4)})}}{x^3}\end{align*}$

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Example Question #2898 : Calculus 3

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

Possible Answers:

${\frac{-{(cos{(y^2)}^3e^{(9z)})}}{{(3x^4)}}}$

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

${\frac{{(cos{(y^2)}^3ln{(3x)}e^{(9z)})}}{{(9x^2)}}}$

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

Correct answer:

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

Explanation:

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

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #2889 : Calculus 3

Example Question #2890 : Calculus 3

Example Question #2891 : Calculus 3

Example Question #2892 : Calculus 3

Example Question #2893 : Calculus 3

Example Question #2894 : Calculus 3

Example Question #2895 : Calculus 3

Example Question #2896 : Calculus 3

Example Question #2897 : Calculus 3

Example Question #2898 : Calculus 3

All Calculus 3 Resources

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