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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #591 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yy}\\&\text{Where }f(x,y,z)=5\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}}y^{2} + 5cos{(y^{2})}cos{(z^{3})}e^{{({x)}}}\end{align*}$

Possible Answers:

${20\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}}y - 20ycos{(z^{3})}sin{(y^{2})}e^{{({x)}}}}$

${10\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}} - 10cos{(z^{3})}sin{(y^{2})}e^{{({x)}}} - 20y^{2}cos{(y^{2})}cos{(z^{3})}e^{{({x)}}}}$

${-{(10\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}}y - 10ycos{(z^{3})}sin{(y^{2})}e^{{({x)}}})}\cdot{(10cos{(z^{3})}sin{(y^{2})}e^{{({x)}}} - 10\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}} + 20y^{2}cos{(y^{2})}cos{(z^{3})}e^{{({x)}}})}}$

${{(10\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}}y - 10ycos{(z^{3})}sin{(y^{2})}e^{{({x)}}})}^{2}}$

Correct answer:

${10\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}} - 10cos{(z^{3})}sin{(y^{2})}e^{{({x)}}} - 20y^{2}cos{(y^{2})}cos{(z^{3})}e^{{({x)}}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=5\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}}y^{2} + 5cos{(y^{2})}cos{(z^{3})}e^{{({x)}}}\\&f_{y}=10\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}}y - 10ycos{(z^{3})}sin{(y^{2})}e^{{({x)}}}\\&f_{yy}=10\cdot2^{{({z^{2})}}}\cdot3^{{({4x)}}} - 10cos{(z^{3})}sin{(y^{2})}e^{{({x)}}} - 20y^{2}cos{(y^{2})}cos{(z^{3})}e^{{({x)}}}\end{align*}$

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Example Question #592 : Partial Derivatives

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

Possible Answers:

${{(10ye^{{({z^{2})}}} +\frac{ {(2x^{2}sin{(3y)})}}{z^{2}})}^{2}}$

${20ye^{{({z^{2})}}} +\frac{ {(4x^{2}sin{(3y)})}}{z^{2}}}$

${{(10e^{{({z^{2})}}} +\frac{ {(6x^{2}cos{(3y)})}}{z^{2}})}\cdot{(10ye^{{({z^{2})}}} +\frac{ {(2x^{2}sin{(3y)})}}{z^{2}})}}$

${10e^{{({z^{2})}}} +\frac{ {(6x^{2}cos{(3y)})}}{z^{2}}}$

Correct answer:

${10e^{{({z^{2})}}} +\frac{ {(6x^{2}cos{(3y)})}}{z^{2}}}$

Explanation:

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

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Example Question #593 : Partial Derivatives

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

Possible Answers:

${4z^{3}cos{(y)} + \frac{{(8\cdot2^{{({4z)}}}ycos{(y^{2})}e^{{({3x)}}})}}{3}}$

${-{(2z^{3}cos{(y)} + \frac{{(4\cdot2^{{({4z)}}}ycos{(y^{2})}e^{{({3x)}}})}}{3})}\cdot{(2z^{3}sin{(y)} - \frac{{(4\cdot2^{{({4z)}}}cos{(y^{2})}e^{{({3x)}}})}}{3 }+ \frac{{(8\cdot2^{{({4z)}}}y^{2}sin{(y^{2})}e^{{({3x)}}})}}{3})}}$

${{(2z^{3}cos{(y)} + \frac{{(4\cdot2^{{({4z)}}}ycos{(y^{2})}e^{{({3x)}}})}}{3})}^{2}}$

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

Correct answer:

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

Explanation:

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

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Example Question #594 : Partial Derivatives

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

Possible Answers:

${-{(\frac{{(3sin{(y^{2})}ln{(z)})}}{x^{4} }+ cos{(4z)}e^{{({4x)}}}cos{(y)})}\cdot{(cos{(4z)}e^{{({4x)}}}sin{(y)} - \frac{{(6ycos{(y^{2})}ln{(z)})}}{x^{4}})}}$

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #595 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #596 : Partial Derivatives

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

Possible Answers:

${\frac{{(24xz^{3}e^{{({x^{2})}}})}}{y^{2} }- 5y^{2}e^{{({x)}}}cos{(z)}}$

${\frac{{(12z^{3}e^{{({x^{2})}}})}}{y^{2} }- 5y^{2}e^{{({x)}}}sin{(z)} - 5y^{2}e^{{({x)}}}cos{(z)} + \frac{{(6xz^{4}e^{{({x^{2})}}})}}{y^{2}}}$

${-{(5y^{2}e^{{({x)}}}sin{(z)} - \frac{{(6xz^{4}e^{{({x^{2})}}})}}{y^{2}})}\cdot{(\frac{{(12z^{3}e^{{({x^{2})}}})}}{y^{2} }- 5y^{2}e^{{({x)}}}cos{(z)})}}$

${-{(5y^{2}e^{{({x)}}}cos{(z)} - \frac{{(24xz^{3}e^{{({x^{2})}}})}}{y^{2}})}\cdot{(\frac{{(12z^{3}e^{{({x^{2})}}})}}{y^{2} }- 5y^{2}e^{{({x)}}}cos{(z)})}}$

Correct answer:

${\frac{{(24xz^{3}e^{{({x^{2})}}})}}{y^{2} }- 5y^{2}e^{{({x)}}}cos{(z)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(3z^{4}e^{{({x^{2})}}})}}{y^{2} }- 5y^{2}e^{{({x)}}}sin{(z)}\\&f_{z}=\frac{{(12z^{3}e^{{({x^{2})}}})}}{y^{2} }- 5y^{2}e^{{({x)}}}cos{(z)}\\&f_{zx}=\frac{{(24xz^{3}e^{{({x^{2})}}})}}{y^{2} }- 5y^{2}e^{{({x)}}}cos{(z)}\end{align*}$

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Example Question #597 : Partial Derivatives

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

Possible Answers:

${\frac{{(2ln{(3z)}e^{{({4y)}}})}}{{(5x^{2})} }- \frac{{(8ln{(3z)}e^{{({4y)}}})}}{{(5x)} }+ \frac{{(15x^{2}ln{(z)})}}{y }- \frac{{(5x^{3}ln{(z)})}}{y^{2}}}$

${{(\frac{{(2ln{(3z)}e^{{({4y)}}})}}{{(5x^{2})} }+ \frac{{(15x^{2}ln{(z)})}}{y})}\cdot{(\frac{{(8ln{(3z)}e^{{({4y)}}})}}{{(5x^{2})} }- \frac{{(15x^{2}ln{(z)})}}{y^{2}})}}$

${-{(\frac{{(8ln{(3z)}e^{{({4y)}}})}}{{(5x)} }+ \frac{{(5x^{3}ln{(z)})}}{y^{2}})}\cdot{(\frac{{(2ln{(3z)}e^{{({4y)}}})}}{{(5x^{2})} }+ \frac{{(15x^{2}ln{(z)})}}{y})}}$

${\frac{{(8ln{(3z)}e^{{({4y)}}})}}{{(5x^{2})} }- \frac{{(15x^{2}ln{(z)})}}{y^{2}}}$

Correct answer:

${\frac{{(8ln{(3z)}e^{{({4y)}}})}}{{(5x^{2})} }- \frac{{(15x^{2}ln{(z)})}}{y^{2}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(5x^{3}ln{(z)})}}{y }- \frac{{(2ln{(3z)}e^{{({4y)}}})}}{{(5x)}}\\&f_{x}=\frac{{(2ln{(3z)}e^{{({4y)}}})}}{{(5x^{2})} }+ \frac{{(15x^{2}ln{(z)})}}{y}\\&f_{xy}=\frac{{(8ln{(3z)}e^{{({4y)}}})}}{{(5x^{2})} }- \frac{{(15x^{2}ln{(z)})}}{y^{2}}\end{align*}$

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Example Question #598 : Partial Derivatives

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

Possible Answers:

${\frac{{(4xyln{(3z)})}}{5 }+ \frac{{(6x^{2})}}{{(yz^{3})}}}$

${\frac{{(2x^{3})}}{{(yz^{3})} }+ \frac{{(2xy^{2}ln{(3z)})}}{5 }+ \frac{{(2x^{2}yln{(3z)})}}{5 }+ \frac{{(6x^{2}ln{(y)})}}{z^{3}}}$

${{(\frac{{(2xy^{2}ln{(3z)})}}{5 }+ \frac{{(6x^{2}ln{(y)})}}{z^{3}})}\cdot{(\frac{{(4xyln{(3z)})}}{5 }+ \frac{{(6x^{2})}}{{(yz^{3})})}}}$

${{(\frac{{(2x^{3})}}{{(yz^{3})} }+ \frac{{(2x^{2}yln{(3z)})}}{5})}\cdot{(\frac{{(2xy^{2}ln{(3z)})}}{5 }+ \frac{{(6x^{2}ln{(y)})}}{z^{3}})}}$

Correct answer:

${\frac{{(4xyln{(3z)})}}{5 }+ \frac{{(6x^{2})}}{{(yz^{3})}}}$

Explanation:

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

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Example Question #599 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #600 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zy}\\&\text{Where }f(x,y,z)=3cos{(z)}ln{(x)}ln{(y)} - \frac{{(2\cdot3^{{({3y)}}}ln{(3x)}e^{{({z^{2})}}})}}{5}\end{align*}$

Possible Answers:

${-{(3ln{(x)}ln{(y)}sin{(z)} + \frac{{(4\cdot3^{{({3y)}}}zln{(3x)}e^{{({z^{2})}}})}}{5})}\cdot{(\frac{{(3cos{(z)}ln{(x)})}}{y }- \frac{{(6\cdot3^{{({3y)}}}ln{(3x)}e^{{({z^{2})}}}ln{(3)})}}{5})}}$

${{(3ln{(x)}ln{(y)}sin{(z)} + \frac{{(4\cdot3^{{({3y)}}}zln{(3x)}e^{{({z^{2})}}})}}{5})}\cdot{(\frac{{(3ln{(x)}sin{(z)})}}{y }+ \frac{{(12\cdot3^{{({3y)}}}zln{(3x)}e^{{({z^{2})}}}ln{(3)})}}{5})}}$

${\frac{{(3cos{(z)}ln{(x)})}}{y }- 3ln{(x)}ln{(y)}sin{(z)} - \frac{{(6\cdot3^{{({3y)}}}ln{(3x)}e^{{({z^{2})}}}ln{(3)})}}{5 }- \frac{{(4\cdot3^{{({3y)}}}zln{(3x)}e^{{({z^{2})}}})}}{5}}$

${- \frac{{(3ln{(x)}sin{(z)})}}{y }- \frac{{(12\cdot3^{{({3y)}}}zln{(3x)}e^{{({z^{2})}}}ln{(3)})}}{5}}$

Correct answer:

${- \frac{{(3ln{(x)}sin{(z)})}}{y }- \frac{{(12\cdot3^{{({3y)}}}zln{(3x)}e^{{({z^{2})}}}ln{(3)})}}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=3cos{(z)}ln{(x)}ln{(y)} - \frac{{(2\cdot3^{{({3y)}}}ln{(3x)}e^{{({z^{2})}}})}}{5}\\&f_{z}=- 3ln{(x)}ln{(y)}sin{(z)} - \frac{{(4\cdot3^{{({3y)}}}zln{(3x)}e^{{({z^{2})}}})}}{5}\\&f_{zy}=- \frac{{(3ln{(x)}sin{(z)})}}{y }- \frac{{(12\cdot3^{{({3y)}}}zln{(3x)}e^{{({z^{2})}}}ln{(3)})}}{5}\end{align*}$

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Calculus 3 : Partial Derivatives

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Example Question #591 : Partial Derivatives

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