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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #591 : 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:

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

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

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

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

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 #592 : 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:

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

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

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

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

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 #593 : 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:

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

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

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

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

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 #591 : 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:

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

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

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

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

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 #592 : 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:

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

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

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

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

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 #596 : 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:

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

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

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

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

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 #597 : 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:

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

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

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

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

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 #598 : 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:

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

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

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

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

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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Example Question #601 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #602 : Partial Derivatives

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

Possible Answers:

${15x^{2}zsin{(y^{3})} - 4cos{(4y)}sin{(x^{2})}sin{(4z)} - 2xcos{(x^{2})}sin{(4y)}sin{(4z)} + 15x^{3}y^{2}zcos{(y^{3})}}$

${45x^{2}y^{2}zcos{(y^{3})} - 8xcos{(x^{2})}cos{(4y)}sin{(4z)}}$

${-{(15x^{2}zsin{(y^{3})} - 2xcos{(x^{2})}sin{(4y)}sin{(4z)})}\cdot{(8xcos{(x^{2})}cos{(4y)}sin{(4z)} - 45x^{2}y^{2}zcos{(y^{3})})}}$

${-{(15x^{2}zsin{(y^{3})} - 2xcos{(x^{2})}sin{(4y)}sin{(4z)})}\cdot{(4cos{(4y)}sin{(x^{2})}sin{(4z)} - 15x^{3}y^{2}zcos{(y^{3})})}}$

Correct answer:

${45x^{2}y^{2}zcos{(y^{3})} - 8xcos{(x^{2})}cos{(4y)}sin{(4z)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=5x^{3}zsin{(y^{3})} - sin{(x^{2})}sin{(4y)}sin{(4z)}\\&f_{x}=15x^{2}zsin{(y^{3})} - 2xcos{(x^{2})}sin{(4y)}sin{(4z)}\\&f_{xy}=45x^{2}y^{2}zcos{(y^{3})} - 8xcos{(x^{2})}cos{(4y)}sin{(4z)}\end{align*}$

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Calculus 3 : Calculus 3

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Example Question #591 : Partial Derivatives

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