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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #621 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=2ln(z^{2})e^{(y^{2})}ln(x) -\frac{ (4^xcos(4y + y^{2} - 1)cos(3z + z^{2} + 1))}{4}\end{align*}$

Possible Answers:

${\frac{(2ln(z^{2})e^{(y^{2})})}{x}+ 4yln(z^{2})e^{(y^{2})}ln(x) +\frac{ (4^xcos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4}-\frac{ (4^xln(4)cos(4y + y^{2} - 1)cos(3z + z^{2} + 1))}{4}}$

${(4yln(z^{2})e^{(y^{2})}ln(x) +\frac{ (4^xcos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4})\cdot(\frac{(2ln(z^{2})e^{(y^{2})})}{x}-\frac{ (4^xln(4)cos(4y + y^{2} - 1)cos(3z + z^{2} + 1))}{4})}$

${\frac{(4yln(z^{2})e^{(y^{2})})}{x}+\frac{ (4^xln(4)cos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4}}$

${(4yln(z^{2})e^{(y^{2})}ln(x) +\frac{ (4^xcos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4})\cdot(\frac{(4yln(z^{2})e^{(y^{2})})}{x}+\frac{ (4^xln(4)cos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4})}$

Correct answer:

${\frac{(4yln(z^{2})e^{(y^{2})})}{x}+\frac{ (4^xln(4)cos(3z + z^{2} + 1)sin(4y + y^{2} - 1)\cdot(2y + 4))}{4}}$

Explanation:

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

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Example Question #621 : Partial Derivatives

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

Possible Answers:

${\frac{(30y^{2}cos(z^{3})sin(y^{3}))}{x^{3}}-\frac{ (ln(z)sin(x))}{(4y)}}$

${(\frac{(ln(z)sin(x))}{(4y)}-\frac{ (30y^{2}cos(z^{3})sin(y^{3}))}{x^{3}})\cdot(\frac{(ln(4y)ln(z)sin(x))}{4}+\frac{ (10cos(y^{3})cos(z^{3}))}{x^{3}})}$

${-(\frac{(ln(4y)ln(z)sin(x))}{4}+\frac{ (10cos(y^{3})cos(z^{3}))}{x^{3}})\cdot(\frac{(cos(x)ln(z))}{(4y)}-\frac{ (15y^{2}cos(z^{3})sin(y^{3}))}{x^{2}})}$

${\frac{(cos(x)ln(z))}{(4y)}-\frac{ (10cos(y^{3})cos(z^{3}))}{x^{3}}-\frac{ (ln(4y)ln(z)sin(x))}{4}-\frac{ (15y^{2}cos(z^{3})sin(y^{3}))}{x^{2}}}$

Correct answer:

${\frac{(30y^{2}cos(z^{3})sin(y^{3}))}{x^{3}}-\frac{ (ln(z)sin(x))}{(4y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(ln(4y)cos(x)ln(z))}{4}+\frac{ (5cos(y^{3})cos(z^{3}))}{x^{2}}\\&f_{x}=-\frac{ (ln(4y)ln(z)sin(x))}{4}-\frac{ (10cos(y^{3})cos(z^{3}))}{x^{3}}\\&f_{xy}=\frac{(30y^{2}cos(z^{3})sin(y^{3}))}{x^{3}}-\frac{ (ln(z)sin(x))}{(4y)}\end{align*}$

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Example Question #2991 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zx}\\&\text{Where }f(x,y,z)=\frac{(5z^{2}sin(y^{2}))}{x^{2}}+\frac{ (2e^{(x^{2})}e^{(3z)}ln(y))}{7}\end{align*}$

Possible Answers:

${\frac{(6e^{(x^{2})}e^{(3z)}ln(y))}{7}-\frac{ (10z^{2}sin(y^{2}))}{x^{3}}+\frac{ (10zsin(y^{2}))}{x^{2}}+\frac{ (4xe^{(x^{2})}e^{(3z)}ln(y))}{7}}$

${-(\frac{(6e^{(x^{2})}e^{(3z)}ln(y))}{7}+\frac{ (10zsin(y^{2}))}{x^{2}})\cdot(\frac{(20zsin(y^{2}))}{x^{3}}-\frac{ (12xe^{(x^{2})}e^{(3z)}ln(y))}{7})}$

${-(\frac{(6e^{(x^{2})}e^{(3z)}ln(y))}{7}+\frac{ (10zsin(y^{2}))}{x^{2}})\cdot(\frac{(10z^{2}sin(y^{2}))}{x^{3}}-\frac{ (4xe^{(x^{2})}e^{(3z)}ln(y))}{7})}$

${\frac{(12xe^{(x^{2})}e^{(3z)}ln(y))}{7}-\frac{ (20zsin(y^{2}))}{x^{3}}}$

Correct answer:

${\frac{(12xe^{(x^{2})}e^{(3z)}ln(y))}{7}-\frac{ (20zsin(y^{2}))}{x^{3}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(5z^{2}sin(y^{2}))}{x^{2}}+\frac{ (2e^{(x^{2})}e^{(3z)}ln(y))}{7}\\&f_{z}=\frac{(6e^{(x^{2})}e^{(3z)}ln(y))}{7}+\frac{ (10zsin(y^{2}))}{x^{2}}\\&f_{zx}=\frac{(12xe^{(x^{2})}e^{(3z)}ln(y))}{7}-\frac{ (20zsin(y^{2}))}{x^{3}}\end{align*}$

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Example Question #621 : Partial Derivatives

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

Possible Answers:

${- 8cos(y^{2})e^{(4z)}sin(3x + x^{2}) -\frac{ (4zsin(z^{2})sin(y))}{x^{2}}}$

${- 16cos(y^{2})e^{(4z)}sin(3x + x^{2}) -\frac{ (2sin(z^{2})sin(y))}{x^{2}}-\frac{ (4z^{2}cos(z^{2})sin(y))}{x^{2}}}$

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

${(4cos(y^{2})e^{(4z)}sin(3x + x^{2}) +\frac{ (2zsin(z^{2})sin(y))}{x^{2}})\cdot(16cos(y^{2})e^{(4z)}sin(3x + x^{2}) +\frac{ (2sin(z^{2})sin(y))}{x^{2}}+\frac{ (4z^{2}cos(z^{2})sin(y))}{x^{2}})}$

Correct answer:

${- 16cos(y^{2})e^{(4z)}sin(3x + x^{2}) -\frac{ (2sin(z^{2})sin(y))}{x^{2}}-\frac{ (4z^{2}cos(z^{2})sin(y))}{x^{2}}}$

Explanation:

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

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Example Question #2993 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #2994 : Calculus 3

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

Possible Answers:

${- 2^{(4y)}\cdot4^{(x^{2})}ln(3z)ln(4) - 2\cdot2^{(4y)}\cdot4^{(x^{2})}x^{2}ln(3z)ln(4)^{2}}$

${-2\cdot2^{(4y)}\cdot4^{(x^{2})}xln(3z)ln(4)}$

${2^{(8y)}\cdot4^{(2x^{2})}x^{2}ln(3z)^{2}ln(4)^{2}}$

${2^{(4y)}\cdot4^{(x^{2})}xln(3z)ln(4)\cdot(2^{(4y)}\cdot4^{(x^{2})}ln(3z)ln(4) + 2\cdot2^{(4y)}\cdot4^{(x^{2})}x^{2}ln(3z)ln(4)^{2})}$

Correct answer:

${- 2^{(4y)}\cdot4^{(x^{2})}ln(3z)ln(4) - 2\cdot2^{(4y)}\cdot4^{(x^{2})}x^{2}ln(3z)ln(4)^{2}}$

Explanation:

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

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Example Question #2995 : Calculus 3

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

Possible Answers:

${\frac{(4\cdot3^{(4x)}cos(y^{2})ln(3))}{(5z)}}$

${\frac{(4\cdot3^{(4x)}cos(y^{2})ln(3)\cdot(2y^{4}cos(z) +\frac{ (3^{(4x)}cos(y^{2}))}{(5z)}))}{(5z)}}$

${2y^{4}cos(z) +\frac{ (3^{(4x)}cos(y^{2}))}{(5z)}+\frac{ (4\cdot3^{(4x)}cos(y^{2})ln(3)ln(z))}{5}}$

${\frac{(4\cdot3^{(4x)}cos(y^{2})ln(3)ln(z)\cdot(2y^{4}cos(z) +\frac{ (3^{(4x)}cos(y^{2}))}{(5z)}))}{5}}$

Correct answer:

${\frac{(4\cdot3^{(4x)}cos(y^{2})ln(3))}{(5z)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=2y^{4}sin(z) +\frac{ (3^{(4x)}cos(y^{2})ln(z))}{5}\\&f_{z}=2y^{4}cos(z) +\frac{ (3^{(4x)}cos(y^{2}))}{(5z)}\\&f_{zx}=\frac{(4\cdot3^{(4x)}cos(y^{2})ln(3))}{(5z)}\end{align*}$

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Example Question #2996 : Calculus 3

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

Possible Answers:

${8\cdot2^{(4x)}ln(3z)ln(2)sin(y^{2} - 4y)\cdot(2y - 4)}$

${2\cdot2^{(4x)}ln(3z)sin(y^{2} - 4y)\cdot(2y - 4) - 8\cdot2^{(4x)}ln(3z)ln(2)cos(y^{2} - 4y)}$

${16\cdot2^{(8x)}ln(3z)^{2}ln(2)sin(y^{2} - 4y)^{2}\cdot(2y - 4)^{2}}$

${-16\cdot2^{(8x)}ln(3z)^{2}ln(2)cos(y^{2} - 4y)sin(y^{2} - 4y)\cdot(2y - 4)}$

Correct answer:

${8\cdot2^{(4x)}ln(3z)ln(2)sin(y^{2} - 4y)\cdot(2y - 4)}$

Explanation:

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

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Example Question #2997 : Calculus 3

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

Possible Answers:

${4cos(y^{2})ln(x)sin(z) - 36e^{(3y)}cos(x^{2} - 3x)cos(z) - 8y^{2}sin(y^{2})ln(x)sin(z)}$

${8ycos(y^{2})ln(x)sin(z) - 24e^{(3y)}cos(x^{2} - 3x)cos(z)}$

${(12e^{(3y)}cos(x^{2} - 3x)cos(z) - 4ycos(y^{2})ln(x)sin(z))\cdot(36e^{(3y)}cos(x^{2} - 3x)cos(z) - 4cos(y^{2})ln(x)sin(z) + 8y^{2}sin(y^{2})ln(x)sin(z))}$

${(12e^{(3y)}cos(x^{2} - 3x)cos(z) - 4ycos(y^{2})ln(x)sin(z))^{2}}$

Correct answer:

${4cos(y^{2})ln(x)sin(z) - 36e^{(3y)}cos(x^{2} - 3x)cos(z) - 8y^{2}sin(y^{2})ln(x)sin(z)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=2sin(y^{2})ln(x)sin(z) - 4e^{(3y)}cos(x^{2} - 3x)cos(z)\\&f_{y}=4ycos(y^{2})ln(x)sin(z) - 12e^{(3y)}cos(x^{2} - 3x)cos(z)\\&f_{yy}=4cos(y^{2})ln(x)sin(z) - 36e^{(3y)}cos(x^{2} - 3x)cos(z) - 8y^{2}sin(y^{2})ln(x)sin(z)\end{align*}$

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Example Question #2998 : Calculus 3

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

Possible Answers:

${\frac{1}{(xyz)}-\frac{ (12x^{2}cos(y))}{z}}$

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

${(\frac{1}{(xyz)}-\frac{ (12x^{2}cos(y))}{z})\cdot(\frac{ln(4z)}{(xy)}- 12x^{2}cos(y)ln(z))}$

${(\frac{ln(4x)}{(yz)}-\frac{ (4x^{3}cos(y))}{z})\cdot(\frac{ln(4z)}{(xy)}- 12x^{2}cos(y)ln(z))}$

Correct answer:

${\frac{1}{(xyz)}-\frac{ (12x^{2}cos(y))}{z}}$

Explanation:

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #621 : Partial Derivatives

Example Question #621 : Partial Derivatives

Example Question #2991 : Calculus 3

Example Question #621 : Partial Derivatives

Example Question #2993 : Calculus 3

Example Question #2994 : Calculus 3

Example Question #2995 : Calculus 3

Example Question #2996 : Calculus 3

Example Question #2997 : Calculus 3

Example Question #2998 : Calculus 3

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