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Calculus 3 : Calculus 3

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Example Questions

Example Question #3131 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=-\frac{ (2sin(x^{2})sin(4y + y^{2})tan(11z + z^{4}))}{5}-\frac{ (2x^{4}z^{2})}{y}\end{align*}$

Possible Answers:

${\frac{(2x^{4}z^{2})}{y^{2}}-\frac{ (8x^{3}z^{2})}{y}-\frac{ (4xcos(x^{2})sin(4y + y^{2})tan(11z + z^{4}))}{5}-\frac{ (2sin(x^{2})cos(4y + y^{2})tan(11z + z^{4})\cdot (2y + 4))}{5}}$

${\frac{(8x^{3}z^{2})}{y^{2}}-\frac{ (4xcos(x^{2})cos(4y + y^{2})tan(11z + z^{4})\cdot (2y + 4))}{5}}$

${-(\frac{(2x^{4}z^{2})}{y^{2}}-\frac{ (2sin(x^{2})cos(4y + y^{2})tan(11z + z^{4})\cdot (2y + 4))}{5})\cdot (\frac{(8x^{3}z^{2})}{y}+\frac{ (4xcos(x^{2})sin(4y + y^{2})tan(11z + z^{4}))}{5})}$

${(\frac{(2x^{4}z^{2})}{y^{2}}-\frac{ (2sin(x^{2})cos(4y + y^{2})tan(11z + z^{4})\cdot (2y + 4))}{5})\cdot (\frac{(8x^{3}z^{2})}{y^{2}}-\frac{ (4xcos(x^{2})cos(4y + y^{2})tan(11z + z^{4})\cdot (2y + 4))}{5})}$

Correct answer:

${\frac{(8x^{3}z^{2})}{y^{2}}-\frac{ (4xcos(x^{2})cos(4y + y^{2})tan(11z + z^{4})\cdot (2y + 4))}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{ (2sin(x^{2})sin(4y + y^{2})tan(11z + z^{4}))}{5}-\frac{ (2x^{4}z^{2})}{y}\\&f_{y}=\frac{(2x^{4}z^{2})}{y^{2}}-\frac{ (2sin(x^{2})cos(4y + y^{2})tan(11z + z^{4})\cdot (2y + 4))}{5}\\&f_{yx}=\frac{(8x^{3}z^{2})}{y^{2}}-\frac{ (4xcos(x^{2})cos(4y + y^{2})tan(11z + z^{4})\cdot (2y + 4))}{5}\end{align*}$

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Example Question #3132 : Calculus 3

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

Possible Answers:

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

${-(\frac{(8y^{4})}{(x^{3}z^{3})}-\frac{ (4xz^{2}sin(x^{2})e^{(y^{2})})}{5})\cdot (\frac{(4z^{2}sin(x^{2})e^{(y^{2})})}{5}+\frac{ (24y^{4})}{(x^{4}z^{3})}+\frac{ (8x^{2}z^{2}cos(x^{2})e^{(y^{2})})}{5})}$

${-\frac{ (4z^{2}sin(x^{2})e^{(y^{2})})}{5}-\frac{ (24y^{4})}{(x^{4}z^{3})}-\frac{ (8x^{2}z^{2}cos(x^{2})e^{(y^{2})})}{5}}$

${\frac{(16y^{4})}{(x^{3}z^{3})}-\frac{ (8xz^{2}sin(x^{2})e^{(y^{2})})}{5}}$

Correct answer:

${-\frac{ (4z^{2}sin(x^{2})e^{(y^{2})})}{5}-\frac{ (24y^{4})}{(x^{4}z^{3})}-\frac{ (8x^{2}z^{2}cos(x^{2})e^{(y^{2})})}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2z^{2}cos(x^{2})e^{(y^{2})})}{5}-\frac{ (4y^{4})}{(x^{2}z^{3})}\\&f_{x}=\frac{(8y^{4})}{(x^{3}z^{3})}-\frac{ (4xz^{2}sin(x^{2})e^{(y^{2})})}{5}\\&f_{xx}=-\frac{ (4z^{2}sin(x^{2})e^{(y^{2})})}{5}-\frac{ (24y^{4})}{(x^{4}z^{3})}-\frac{ (8x^{2}z^{2}cos(x^{2})e^{(y^{2})})}{5}\end{align*}$

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Example Question #3133 : Calculus 3

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

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Example Question #3134 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yy}\\&\text{Where }f(x,y,z)=5sin(y^{2})sin(z) +\frac{ (cos(x^{3} - 3x)tan(12y + y^{4})sin(4z + z^{2}))}{2}\end{align*}$

Possible Answers:

${(10ycos(y^{2})sin(z) +\frac{ (cos(x^{3} - 3x)sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1)\cdot (4y^{3} + 12))}{2})^{2}}$

${20ycos(y^{2})sin(z) + cos(x^{3} - 3x)sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1)\cdot (4y^{3} + 12)}$

${(10ycos(y^{2})sin(z) +\frac{ (cos(x^{3} - 3x)sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1)\cdot (4y^{3} + 12))}{2})\cdot (10cos(y^{2})sin(z) - 20y^{2}sin(y^{2})sin(z) + 6y^{2}cos(x^{3} - 3x)sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1) + cos(x^{3} - 3x)tan(12y + y^{4})sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1)\cdot (4y^{3} + 12)^{2})}$

${10cos(y^{2})sin(z) - 20y^{2}sin(y^{2})sin(z) + 6y^{2}cos(x^{3} - 3x)sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1) + cos(x^{3} - 3x)tan(12y + y^{4})sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1)\cdot (4y^{3} + 12)^{2}}$

Correct answer:

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=5sin(y^{2})sin(z) +\frac{ (cos(x^{3} - 3x)tan(12y + y^{4})sin(4z + z^{2}))}{2}\\&f_{y}=10ycos(y^{2})sin(z) +\frac{ (cos(x^{3} - 3x)sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1)\cdot (4y^{3} + 12))}{2}\\&f_{yy}=10cos(y^{2})sin(z) - 20y^{2}sin(y^{2})sin(z) + 6y^{2}cos(x^{3} - 3x)sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1) + cos(x^{3} - 3x)tan(12y + y^{4})sin(4z + z^{2})\cdot (tan(12y + y^{4})^{2} + 1)\cdot (4y^{3} + 12)^{2}\end{align*}$

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Example Question #3135 : Calculus 3

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

Possible Answers:

${-(\frac{(x^{2}e^{(y^{2})})}{z}- 8x^{4}zcos(z^{2})ln(y))\cdot (\frac{(8x^{4}zcos(z^{2}))}{y}-\frac{ (2x^{2}ye^{(y^{2})})}{z})}$

${-(\frac{(x^{2}e^{(y^{2})})}{z}- 8x^{4}zcos(z^{2})ln(y))\cdot (\frac{(4x^{4}sin(z^{2}))}{y}- x^{2}yln(z^{2})e^{(y^{2})})}$

${\frac{(4x^{4}sin(z^{2}))}{y}-\frac{ (x^{2}e^{(y^{2})})}{z}- x^{2}yln(z^{2})e^{(y^{2})} + 8x^{4}zcos(z^{2})ln(y)}$

${\frac{(8x^{4}zcos(z^{2}))}{y}-\frac{ (2x^{2}ye^{(y^{2})})}{z}}$

Correct answer:

${\frac{(8x^{4}zcos(z^{2}))}{y}-\frac{ (2x^{2}ye^{(y^{2})})}{z}}$

Explanation:

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Example Question #3136 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{zx}\\&\text{Where }f(x,y,z)=-2sin(y^{2})tan(12x + x^{4})tan(11z + z^{2})\end{align*}$

Possible Answers:

${- 2sin(y^{2})tan(11z + z^{2})\cdot (tan(12x + x^{4})^{2} + 1)\cdot (4x^{3} + 12) - 2sin(y^{2})tan(12x + x^{4})\cdot (2z + 11)\cdot (tan(11z + z^{2})^{2} + 1)}$

${4sin(y^{2})^{2}tan(12x + x^{4})\cdot (2z + 11)^{2}\cdot (tan(12x + x^{4})^{2} + 1)\cdot (tan(11z + z^{2})^{2} + 1)^{2}\cdot (4x^{3} + 12)}$

${-2sin(y^{2})\cdot (2z + 11)\cdot (tan(12x + x^{4})^{2} + 1)\cdot (tan(11z + z^{2})^{2} + 1)\cdot (4x^{3} + 12)}$

${4sin(y^{2})^{2}tan(12x + x^{4})tan(11z + z^{2})\cdot (2z + 11)\cdot (tan(12x + x^{4})^{2} + 1)\cdot (tan(11z + z^{2})^{2} + 1)\cdot (4x^{3} + 12)}$

Correct answer:

${-2sin(y^{2})\cdot (2z + 11)\cdot (tan(12x + x^{4})^{2} + 1)\cdot (tan(11z + z^{2})^{2} + 1)\cdot (4x^{3} + 12)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-2sin(y^{2})tan(12x + x^{4})tan(11z + z^{2})\\&f_{z}=-2sin(y^{2})tan(12x + x^{4})\cdot (2z + 11)\cdot (tan(11z + z^{2})^{2} + 1)\\&f_{zx}=-2sin(y^{2})\cdot (2z + 11)\cdot (tan(12x + x^{4})^{2} + 1)\cdot (tan(11z + z^{2})^{2} + 1)\cdot (4x^{3} + 12)\end{align*}$

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Example Question #3137 : Calculus 3

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

Possible Answers:

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

${\frac{(2e^{(x^{2})}cos(y^{2} - 4y)cos(z^{4} - 4z)\cdot (2y - 4)\cdot (2x^{3}zcos(z^{2}) -\frac{ (2e^{(x^{2})}cos(z^{4} - 4z)sin(y^{2} - 4y)\cdot (4z^{3} - 4))}{9})\cdot (4z^{3} - 4)\cdot (\frac{(8z^{2}e^{(x^{2})}cos(y^{2} - 4y)cos(z^{4} - 4z)\cdot (2y - 4))}{3}-\frac{ (2e^{(x^{2})}cos(y^{2} - 4y)sin(z^{4} - 4z)\cdot (2y - 4)\cdot (4z^{3} - 4)^{2})}{9}))}{9}}$

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

${\frac{(2e^{(x^{2})}cos(y^{2} - 4y)sin(z^{4} - 4z)\cdot (2y - 4)\cdot (4z^{3} - 4)^{2})}{9}-\frac{ (8z^{2}e^{(x^{2})}cos(y^{2} - 4y)cos(z^{4} - 4z)\cdot (2y - 4))}{3}}$

Correct answer:

${\frac{(2e^{(x^{2})}cos(y^{2} - 4y)sin(z^{4} - 4z)\cdot (2y - 4)\cdot (4z^{3} - 4)^{2})}{9}-\frac{ (8z^{2}e^{(x^{2})}cos(y^{2} - 4y)cos(z^{4} - 4z)\cdot (2y - 4))}{3}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=x^{3}sin(z^{2}) -\frac{ (2e^{(x^{2})}sin(y^{2} - 4y)sin(z^{4} - 4z))}{9}\\&f_{z}=2x^{3}zcos(z^{2}) -\frac{ (2e^{(x^{2})}cos(z^{4} - 4z)sin(y^{2} - 4y)\cdot (4z^{3} - 4))}{9}\\&f_{zy}=-\frac{(2e^{(x^{2})}cos(y^{2} - 4y)cos(z^{4} - 4z)\cdot (2y - 4)\cdot (4z^{3} - 4))}{9}\\&f_{zyz}=\frac{(2e^{(x^{2})}cos(y^{2} - 4y)sin(z^{4} - 4z)\cdot (2y - 4)\cdot (4z^{3} - 4)^{2})}{9}-\frac{ (8z^{2}e^{(x^{2})}cos(y^{2} - 4y)cos(z^{4} - 4z)\cdot (2y - 4))}{3}\end{align*}$

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Example Question #3138 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3139 : Calculus 3

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

Possible Answers:

${- 4zsin(x^{2})sin(y) -\frac{ (8z^{3}tan(x^{2})e^{(z^{4})})}{y^{3}}- 2xz^{2}cos(x^{2})sin(y) -\frac{ (2xe^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}}$

${-(2zsin(x^{2})sin(y) +\frac{ (4z^{3}tan(x^{2})e^{(z^{4})})}{y^{3}})^{2}\cdot (2xz^{2}cos(x^{2})sin(y) +\frac{ (2xe^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}})}$

${- 4xcos(x^{2})sin(y) -\frac{ (24xz^{2}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}-\frac{ (32xz^{6}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}}$

${-(2zsin(x^{2})sin(y) +\frac{ (4z^{3}tan(x^{2})e^{(z^{4})})}{y^{3}})\cdot (4xcos(x^{2})sin(y) +\frac{ (24xz^{2}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}+\frac{ (32xz^{6}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}})\cdot (2sin(x^{2})sin(y) +\frac{ (12z^{2}tan(x^{2})e^{(z^{4})})}{y^{3}}+\frac{ (16z^{6}tan(x^{2})e^{(z^{4})})}{y^{3}})}$

Correct answer:

${- 4xcos(x^{2})sin(y) -\frac{ (24xz^{2}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}-\frac{ (32xz^{6}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=- z^{2}sin(x^{2})sin(y) -\frac{ (tan(x^{2})e^{(z^{4})})}{y^{3}}\\&f_{z}=- 2zsin(x^{2})sin(y) -\frac{ (4z^{3}tan(x^{2})e^{(z^{4})})}{y^{3}}\\&f_{zz}=- 2sin(x^{2})sin(y) -\frac{ (12z^{2}tan(x^{2})e^{(z^{4})})}{y^{3}}-\frac{ (16z^{6}tan(x^{2})e^{(z^{4})})}{y^{3}}\\&f_{zzx}=- 4xcos(x^{2})sin(y) -\frac{ (24xz^{2}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}-\frac{ (32xz^{6}e^{(z^{4})}\cdot (tan(x^{2})^{2} + 1))}{y^{3}}\end{align*}$

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Example Question #3140 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{yzz}\\&\text{Where }f(x,y,z)=5x^{4}y^{4}z^{2} - 3z^{2}e^{(3y)}tan(11x + x^{4})\end{align*}$

Possible Answers:

${(20x^{4}y^{3}z^{2} - 9z^{2}e^{(3y)}tan(11x + x^{4}))\cdot (10x^{4}y^{4}z - 6ze^{(3y)}tan(11x + x^{4}))^{2}}$

${(20x^{4}y^{3}z^{2} - 9z^{2}e^{(3y)}tan(11x + x^{4}))\cdot (40x^{4}y^{3}z - 18ze^{(3y)}tan(11x + x^{4}))\cdot (40x^{4}y^{3} - 18e^{(3y)}tan(11x + x^{4}))}$

${40x^{4}y^{3} - 18e^{(3y)}tan(11x + x^{4})}$

${20x^{4}y^{4}z - 12ze^{(3y)}tan(11x + x^{4}) + 20x^{4}y^{3}z^{2} - 9z^{2}e^{(3y)}tan(11x + x^{4})}$

Correct answer:

${40x^{4}y^{3} - 18e^{(3y)}tan(11x + x^{4})}$

Explanation:

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Calculus 3 : Calculus 3

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