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Calculus 3 : Calculus 3

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Example Questions

Example Question #3281 : Calculus 3

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

Possible Answers:

${-24z^{2}tan(z^{3})^{2}cos(4y + y^{4})tan(x^{3} - 11x)^{3}sin(4y + y^{4})^{2}\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)^{2}}$ $\displaystyle {-24z^{2}tan(z^{3})^{2}cos(4y + y^{4})tan(x^{3} - 11x)^{3}sin(4y + y^{4})^{2}\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)^{2}}$

${-36z^{4}cos(4y + y^{4})tan(x^{3} - 11x)^{2}sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1)^{2}\cdot (4y^{3} + 4)\cdot (72y^{2}z^{2}tan(x^{3} - 11x)sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1) + 6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)^{2})}$ $\displaystyle {-36z^{4}cos(4y + y^{4})tan(x^{3} - 11x)^{2}sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1)^{2}\cdot (4y^{3} + 4)\cdot (72y^{2}z^{2}tan(x^{3} - 11x)sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1) + 6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)^{2})}$

${72y^{2}z^{2}tan(x^{3} - 11x)sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1) + 6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)^{2}}$ $\displaystyle {72y^{2}z^{2}tan(x^{3} - 11x)sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1) + 6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)^{2}}$

${4tan(z^{3})tan(x^{3} - 11x)sin(4y + y^{4})\cdot (4y^{3} + 4) - 6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)}$ $\displaystyle {4tan(z^{3})tan(x^{3} - 11x)sin(4y + y^{4})\cdot (4y^{3} + 4) - 6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)}$

Correct answer:

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-2tan(z^{3})cos(4y + y^{4})tan(x^{3} - 11x)\\&f_{z}=-6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)\\&f_{zy}=6z^{2}tan(x^{3} - 11x)sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)\\&f_{zyy}=72y^{2}z^{2}tan(x^{3} - 11x)sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1) + 6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)^{2}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-2tan(z^{3})cos(4y + y^{4})tan(x^{3} - 11x)\\&f_{z}=-6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)\\&f_{zy}=6z^{2}tan(x^{3} - 11x)sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)\\&f_{zyy}=72y^{2}z^{2}tan(x^{3} - 11x)sin(4y + y^{4})\cdot (tan(z^{3})^{2} + 1) + 6z^{2}cos(4y + y^{4})tan(x^{3} - 11x)\cdot (tan(z^{3})^{2} + 1)\cdot (4y^{3} + 4)^{2}\end{align*}$

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Example Question #3282 : Calculus 3

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

Possible Answers:

${\frac{(19683\cdot 2^{(9y)}z^{6}sin(x^{2})^{3}ln(2)^{6})}{8}}$ $\displaystyle {\frac{(19683\cdot 2^{(9y)}z^{6}sin(x^{2})^{3}ln(2)^{6})}{8}}$

${\frac{(81\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2)^{3})}{2}}$ $\displaystyle {\frac{(81\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2)^{3})}{2}}$

${\frac{(27\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2))}{2}}$ $\displaystyle {\frac{(27\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2))}{2}}$

${\frac{(729\cdot 2^{(9y)}z^{6}sin(x^{2})^{3}ln(2)^{3})}{8}}$ $\displaystyle {\frac{(729\cdot 2^{(9y)}z^{6}sin(x^{2})^{3}ln(2)^{3})}{8}}$

Correct answer:

${\frac{(81\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2)^{3})}{2}}$ $\displaystyle {\frac{(81\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2)^{3})}{2}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3\cdot 2^{(3y)}z^{2}sin(x^{2}))}{2}\\&f_{y}=\frac{(9\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2))}{2}\\&f_{yy}=\frac{(27\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2)^{2})}{2}\\&f_{yyy}=\frac{(81\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2)^{3})}{2}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(3\cdot 2^{(3y)}z^{2}sin(x^{2}))}{2}\\&f_{y}=\frac{(9\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2))}{2}\\&f_{yy}=\frac{(27\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2)^{2})}{2}\\&f_{yyy}=\frac{(81\cdot 2^{(3y)}z^{2}sin(x^{2})ln(2)^{3})}{2}\end{align*}$

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Example Question #3283 : Calculus 3

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

Possible Answers:

${\frac{(256x^{6}y^{6})}{z^{4}}}$ $\displaystyle {\frac{(256x^{6}y^{6})}{z^{4}}}$

${\frac{(8xy^{2})}{z}+\frac{ (8x^{2}y)}{z}}$ $\displaystyle {\frac{(8xy^{2})}{z}+\frac{ (8x^{2}y)}{z}}$

${\frac{(1024x^{5}y)}{z^{4}}}$ $\displaystyle {\frac{(1024x^{5}y)}{z^{4}}}$

${\frac{8}{z}}$ $\displaystyle {\frac{8}{z}}$

Correct answer:

${\frac{8}{z}}$ $\displaystyle {\frac{8}{z}}$

Explanation:

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Example Question #3284 : Calculus 3

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

Possible Answers:

${(3y\cdot (tan(x)^{2} + 1) +\frac{ tan(y^{2} - 12y)}{(5xz)})\cdot (3tan(x)^{2} +\frac{ ((2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5xz)}+ 3)}$ $\displaystyle {(3y\cdot (tan(x)^{2} + 1) +\frac{ tan(y^{2} - 12y)}{(5xz)})\cdot (3tan(x)^{2} +\frac{ ((2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5xz)}+ 3)}$

${3tan(x)^{2} +\frac{ ((2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5xz)}+ 3}$ $\displaystyle {3tan(x)^{2} +\frac{ ((2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5xz)}+ 3}$

${3tan(x) + 3y\cdot (tan(x)^{2} + 1) +\frac{ tan(y^{2} - 12y)}{(5xz)}+\frac{ (ln(4x)\cdot (2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5z)}}$ $\displaystyle {3tan(x) + 3y\cdot (tan(x)^{2} + 1) +\frac{ tan(y^{2} - 12y)}{(5xz)}+\frac{ (ln(4x)\cdot (2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5z)}}$

${(3y\cdot (tan(x)^{2} + 1) +\frac{ tan(y^{2} - 12y)}{(5xz)})\cdot (3tan(x) +\frac{ (ln(4x)\cdot (2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5z)})}$ $\displaystyle {(3y\cdot (tan(x)^{2} + 1) +\frac{ tan(y^{2} - 12y)}{(5xz)})\cdot (3tan(x) +\frac{ (ln(4x)\cdot (2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5z)})}$

Correct answer:

${3tan(x)^{2} +\frac{ ((2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5xz)}+ 3}$ $\displaystyle {3tan(x)^{2} +\frac{ ((2y - 12)\cdot (tan(y^{2} - 12y)^{2} + 1))}{(5xz)}+ 3}$

Explanation:

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

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Example Question #3285 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xxxy}\\&\text{Where }f(x,y,z)=-\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)sin(x))}{7}\end{align*}$ $\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{xxxy}\\&\text{Where }f(x,y,z)=-\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)sin(x))}{7}\end{align*}$

Possible Answers:

${-\frac{ (6\cdot 2^{(3z)}tan(y^{2} - 11y)cos(x))}{7}-\frac{ (2\cdot 2^{(3z)}sin(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{7}}$ $\displaystyle {-\frac{ (6\cdot 2^{(3z)}tan(y^{2} - 11y)cos(x))}{7}-\frac{ (2\cdot 2^{(3z)}sin(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{7}}$

${-\frac{(16\cdot 2^{(12z)}tan(y^{2} - 11y)^{3}cos(x)^{3}sin(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{2401}}$ $\displaystyle {-\frac{(16\cdot 2^{(12z)}tan(y^{2} - 11y)^{3}cos(x)^{3}sin(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{2401}}$

${\frac{(16\cdot 2^{(12z)}tan(y^{2} - 11y)^{3}cos(x)^{3}sin(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{2401}}$ $\displaystyle {\frac{(16\cdot 2^{(12z)}tan(y^{2} - 11y)^{3}cos(x)^{3}sin(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{2401}}$

${\frac{(2\cdot 2^{(3z)}cos(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{7}}$ $\displaystyle {\frac{(2\cdot 2^{(3z)}cos(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{7}}$

Correct answer:

${\frac{(2\cdot 2^{(3z)}cos(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{7}}$ $\displaystyle {\frac{(2\cdot 2^{(3z)}cos(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)sin(x))}{7}\\&f_{x}=-\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)cos(x))}{7}\\&f_{xx}=\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)sin(x))}{7}\\&f_{xxx}=\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)cos(x))}{7}\\&f_{xxxy}=\frac{(2\cdot 2^{(3z)}cos(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{7}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)sin(x))}{7}\\&f_{x}=-\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)cos(x))}{7}\\&f_{xx}=\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)sin(x))}{7}\\&f_{xxx}=\frac{(2\cdot 2^{(3z)}tan(y^{2} - 11y)cos(x))}{7}\\&f_{xxxy}=\frac{(2\cdot 2^{(3z)}cos(x)\cdot (2y - 11)\cdot (tan(y^{2} - 11y)^{2} + 1))}{7}\end{align*}$

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Example Question #3286 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #3287 : Calculus 3

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

Possible Answers:

${-\frac{(32768z^{2}tan(z^{3})^{4}e^{(20y)}\cdot (tan(z^{3})^{2} + 1))}{(19683x^{9})}}$ $\displaystyle {-\frac{(32768z^{2}tan(z^{3})^{4}e^{(20y)}\cdot (tan(z^{3})^{2} + 1))}{(19683x^{9})}}$

${\frac{(32z^{2}e^{(4y)}\cdot (tan(z^{3})^{2} + 1))}{(3x^{2})}}$ $\displaystyle {\frac{(32z^{2}e^{(4y)}\cdot (tan(z^{3})^{2} + 1))}{(3x^{2})}}$

${-\frac{(512z^{2}tan(z^{3})^{4}e^{(20y)}ln(x)^{3}\cdot (tan(z^{3})^{2} + 1))}{(19683x^{2})}}$ $\displaystyle {-\frac{(512z^{2}tan(z^{3})^{4}e^{(20y)}ln(x)^{3}\cdot (tan(z^{3})^{2} + 1))}{(19683x^{2})}}$

${-\frac{ (4tan(z^{3})e^{(4y)})}{(9x)}-\frac{ (16tan(z^{3})e^{(4y)}ln(x))}{9}-\frac{ (2z^{2}e^{(4y)}ln(x)\cdot (tan(z^{3})^{2} + 1))}{3}}$ $\displaystyle {-\frac{ (4tan(z^{3})e^{(4y)})}{(9x)}-\frac{ (16tan(z^{3})e^{(4y)}ln(x))}{9}-\frac{ (2z^{2}e^{(4y)}ln(x)\cdot (tan(z^{3})^{2} + 1))}{3}}$

Correct answer:

${\frac{(32z^{2}e^{(4y)}\cdot (tan(z^{3})^{2} + 1))}{(3x^{2})}}$ $\displaystyle {\frac{(32z^{2}e^{(4y)}\cdot (tan(z^{3})^{2} + 1))}{(3x^{2})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{(2tan(z^{3})e^{(4y)}ln(x))}{9}\\&f_{x}=-\frac{(2tan(z^{3})e^{(4y)})}{(9x)}\\&f_{xx}=\frac{(2tan(z^{3})e^{(4y)})}{(9x^{2})}\\&f_{xxy}=\frac{(8tan(z^{3})e^{(4y)})}{(9x^{2})}\\&f_{xxyy}=\frac{(32tan(z^{3})e^{(4y)})}{(9x^{2})}\\&f_{xxyyz}=\frac{(32z^{2}e^{(4y)}\cdot (tan(z^{3})^{2} + 1))}{(3x^{2})}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{(2tan(z^{3})e^{(4y)}ln(x))}{9}\\&f_{x}=-\frac{(2tan(z^{3})e^{(4y)})}{(9x)}\\&f_{xx}=\frac{(2tan(z^{3})e^{(4y)})}{(9x^{2})}\\&f_{xxy}=\frac{(8tan(z^{3})e^{(4y)})}{(9x^{2})}\\&f_{xxyy}=\frac{(32tan(z^{3})e^{(4y)})}{(9x^{2})}\\&f_{xxyyz}=\frac{(32z^{2}e^{(4y)}\cdot (tan(z^{3})^{2} + 1))}{(3x^{2})}\end{align*}$

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Example Question #3288 : Calculus 3

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

Possible Answers:

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

${\frac{(9tan(x^{4})^{2}tan(12y + y^{3})sin(z^{3} - 4z)^{2}\cdot (tan(12y + y^{3})^{2} + 1)\cdot (3y^{2} + 12)\cdot (3z^{2} - 4)^{2})}{4}}$ $\displaystyle {\frac{(9tan(x^{4})^{2}tan(12y + y^{3})sin(z^{3} - 4z)^{2}\cdot (tan(12y + y^{3})^{2} + 1)\cdot (3y^{2} + 12)\cdot (3z^{2} - 4)^{2})}{4}}$

${-\frac{(9tan(x^{4})^{2}cos(z^{3} - 4z)tan(12y + y^{3})sin(z^{3} - 4z)\cdot (tan(12y + y^{3})^{2} + 1)\cdot (3y^{2} + 12)\cdot (3z^{2} - 4))}{4}}$ $\displaystyle {-\frac{(9tan(x^{4})^{2}cos(z^{3} - 4z)tan(12y + y^{3})sin(z^{3} - 4z)\cdot (tan(12y + y^{3})^{2} + 1)\cdot (3y^{2} + 12)\cdot (3z^{2} - 4))}{4}}$

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

Correct answer:

Explanation:

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

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Example Question #3289 : Calculus 3

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

Possible Answers:

${192\cdot 3^{(4z)}xcos(x^{2})cos(y^{2})ln(3)^{2} - 384\cdot 3^{(4z)}xy^{2}cos(x^{2})sin(y^{2})ln(3)^{2}}$ $\displaystyle {192\cdot 3^{(4z)}xcos(x^{2})cos(y^{2})ln(3)^{2} - 384\cdot 3^{(4z)}xy^{2}cos(x^{2})sin(y^{2})ln(3)^{2}}$

${6\cdot 3^{(4z)}xcos(x^{2})sin(y^{2}) + 12\cdot 3^{(4z)}ycos(y^{2})sin(x^{2}) + 24\cdot 3^{(4z)}sin(x^{2})sin(y^{2})ln(3)}$ $\displaystyle {6\cdot 3^{(4z)}xcos(x^{2})sin(y^{2}) + 12\cdot 3^{(4z)}ycos(y^{2})sin(x^{2}) + 24\cdot 3^{(4z)}sin(x^{2})sin(y^{2})ln(3)}$

${31104\cdot 3^{(20z)}xy^{2}cos(x^{2})cos(y^{2})^{2}sin(x^{2})^{4}sin(y^{2})^{3}ln(3)^{2}}$ $\displaystyle {31104\cdot 3^{(20z)}xy^{2}cos(x^{2})cos(y^{2})^{2}sin(x^{2})^{4}sin(y^{2})^{3}ln(3)^{2}}$

${663552\cdot 3^{(16z)}x^{3}y^{4}cos(x^{2})^{3}cos(y^{2})^{4}sin(x^{2})ln(3)^{3}\cdot (192\cdot 3^{(4z)}xcos(x^{2})cos(y^{2})ln(3)^{2} - 384\cdot 3^{(4z)}xy^{2}cos(x^{2})sin(y^{2})ln(3)^{2})}$ $\displaystyle {663552\cdot 3^{(16z)}x^{3}y^{4}cos(x^{2})^{3}cos(y^{2})^{4}sin(x^{2})ln(3)^{3}\cdot (192\cdot 3^{(4z)}xcos(x^{2})cos(y^{2})ln(3)^{2} - 384\cdot 3^{(4z)}xy^{2}cos(x^{2})sin(y^{2})ln(3)^{2})}$

Correct answer:

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3\cdot 3^{(4z)}sin(x^{2})sin(y^{2})\\&f_{y}=6\cdot 3^{(4z)}ycos(y^{2})sin(x^{2})\\&f_{yx}=12\cdot 3^{(4z)}xycos(x^{2})cos(y^{2})\\&f_{yxz}=48\cdot 3^{(4z)}xycos(x^{2})cos(y^{2})ln(3)\\&f_{yxzz}=192\cdot 3^{(4z)}xycos(x^{2})cos(y^{2})ln(3)^{2}\\&f_{yxzzy}=192\cdot 3^{(4z)}xcos(x^{2})cos(y^{2})ln(3)^{2} - 384\cdot 3^{(4z)}xy^{2}cos(x^{2})sin(y^{2})ln(3)^{2}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3\cdot 3^{(4z)}sin(x^{2})sin(y^{2})\\&f_{y}=6\cdot 3^{(4z)}ycos(y^{2})sin(x^{2})\\&f_{yx}=12\cdot 3^{(4z)}xycos(x^{2})cos(y^{2})\\&f_{yxz}=48\cdot 3^{(4z)}xycos(x^{2})cos(y^{2})ln(3)\\&f_{yxzz}=192\cdot 3^{(4z)}xycos(x^{2})cos(y^{2})ln(3)^{2}\\&f_{yxzzy}=192\cdot 3^{(4z)}xcos(x^{2})cos(y^{2})ln(3)^{2} - 384\cdot 3^{(4z)}xy^{2}cos(x^{2})sin(y^{2})ln(3)^{2}\end{align*}$

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Example Question #3289 : Calculus 3

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

Possible Answers:

${\frac{(4\cdot 2^{(6z)}ln(y^{2})^{4}ln(2)^{2})}{(729x^{8}y^{2})}}$ $\displaystyle {\frac{(4\cdot 2^{(6z)}ln(y^{2})^{4}ln(2)^{2})}{(729x^{8}y^{2})}}$

${\frac{(16\cdot 2^{(6z)}ln(y^{2})^{3}ln(2)^{10})}{(729x^{12}y^{5})}}$ $\displaystyle {\frac{(16\cdot 2^{(6z)}ln(y^{2})^{3}ln(2)^{10})}{(729x^{12}y^{5})}}$

${\frac{(2\cdot 2^zln(y^{2}))}{(3x^{2})}-\frac{ (4\cdot 2^z)}{(3xy)}-\frac{ (2\cdot 2^zln(y^{2})ln(2))}{(3x)}}$ $\displaystyle {\frac{(2\cdot 2^zln(y^{2}))}{(3x^{2})}-\frac{ (4\cdot 2^z)}{(3xy)}-\frac{ (2\cdot 2^zln(y^{2})ln(2))}{(3x)}}$

${\frac{(4\cdot 2^zln(2)^{2})}{(3x^{3}y^{2})}}$ $\displaystyle {\frac{(4\cdot 2^zln(2)^{2})}{(3x^{3}y^{2})}}$

Correct answer:

${\frac{(4\cdot 2^zln(2)^{2})}{(3x^{3}y^{2})}}$ $\displaystyle {\frac{(4\cdot 2^zln(2)^{2})}{(3x^{3}y^{2})}}$

Explanation:

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

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Calculus 3 : Calculus 3

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