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Calculus 3 : Partial Derivatives

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Example Questions

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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Example Question #3291 : Calculus 3

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

Possible Answers:

${8ycos(z) - 6\cdot 4^{(y^{2})}yze^{(z^{2})}ln(4)sin(x^{2} - 4x)}$ $\displaystyle {8ycos(z) - 6\cdot 4^{(y^{2})}yze^{(z^{2})}ln(4)sin(x^{2} - 4x)}$

${(4y^{2}cos(z) - 3\cdot 4^{(y^{2})}ze^{(z^{2})}sin(x^{2} - 4x))\cdot (8ysin(z) - 3\cdot 4^{(y^{2})}ye^{(z^{2})}ln(4)sin(x^{2} - 4x))}$ $\displaystyle {(4y^{2}cos(z) - 3\cdot 4^{(y^{2})}ze^{(z^{2})}sin(x^{2} - 4x))\cdot (8ysin(z) - 3\cdot 4^{(y^{2})}ye^{(z^{2})}ln(4)sin(x^{2} - 4x))}$

${4y^{2}cos(z) + 8ysin(z) - 3\cdot 4^{(y^{2})}ze^{(z^{2})}sin(x^{2} - 4x) - 3\cdot 4^{(y^{2})}ye^{(z^{2})}ln(4)sin(x^{2} - 4x)}$ $\displaystyle {4y^{2}cos(z) + 8ysin(z) - 3\cdot 4^{(y^{2})}ze^{(z^{2})}sin(x^{2} - 4x) - 3\cdot 4^{(y^{2})}ye^{(z^{2})}ln(4)sin(x^{2} - 4x)}$

${(4y^{2}cos(z) - 3\cdot 4^{(y^{2})}ze^{(z^{2})}sin(x^{2} - 4x))\cdot (8ycos(z) - 6\cdot 4^{(y^{2})}yze^{(z^{2})}ln(4)sin(x^{2} - 4x))}$ $\displaystyle {(4y^{2}cos(z) - 3\cdot 4^{(y^{2})}ze^{(z^{2})}sin(x^{2} - 4x))\cdot (8ycos(z) - 6\cdot 4^{(y^{2})}yze^{(z^{2})}ln(4)sin(x^{2} - 4x))}$

Correct answer:

${8ycos(z) - 6\cdot 4^{(y^{2})}yze^{(z^{2})}ln(4)sin(x^{2} - 4x)}$ $\displaystyle {8ycos(z) - 6\cdot 4^{(y^{2})}yze^{(z^{2})}ln(4)sin(x^{2} - 4x)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=4y^{2}sin(z) -\frac{ (3\cdot 4^{(y^{2})}e^{(z^{2})}sin(x^{2} - 4x))}{2}\\&f_{z}=4y^{2}cos(z) - 3\cdot 4^{(y^{2})}ze^{(z^{2})}sin(x^{2} - 4x)\\&f_{zy}=8ycos(z) - 6\cdot 4^{(y^{2})}yze^{(z^{2})}ln(4)sin(x^{2} - 4x)\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=4y^{2}sin(z) -\frac{ (3\cdot 4^{(y^{2})}e^{(z^{2})}sin(x^{2} - 4x))}{2}\\&f_{z}=4y^{2}cos(z) - 3\cdot 4^{(y^{2})}ze^{(z^{2})}sin(x^{2} - 4x)\\&f_{zy}=8ycos(z) - 6\cdot 4^{(y^{2})}yze^{(z^{2})}ln(4)sin(x^{2} - 4x)\end{align*}$

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Example Question #3292 : Calculus 3

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

Possible Answers:

${\frac{(3^{(4x)}ln(4y)^{2}ln(3)^{8}tan(z^{3} - 12z)^{4})}{(256y^{2})}}$ $\displaystyle {\frac{(3^{(4x)}ln(4y)^{2}ln(3)^{8}tan(z^{3} - 12z)^{4})}{(256y^{2})}}$

${\frac{(3^xln(3)^{3}tan(z^{3} - 12z))}{(4y)}}$ $\displaystyle {\frac{(3^xln(3)^{3}tan(z^{3} - 12z))}{(4y)}}$

${\frac{(3^xtan(z^{3} - 12z))}{(4y)}+\frac{ (3\cdot 3^xln(4y)ln(3)tan(z^{3} - 12z))}{4}}$ $\displaystyle {\frac{(3^xtan(z^{3} - 12z))}{(4y)}+\frac{ (3\cdot 3^xln(4y)ln(3)tan(z^{3} - 12z))}{4}}$

${\frac{(3^{(4x)}ln(4y)^{3}ln(3)^{3}tan(z^{3} - 12z)^{4})}{(256y)}}$ $\displaystyle {\frac{(3^{(4x)}ln(4y)^{3}ln(3)^{3}tan(z^{3} - 12z)^{4})}{(256y)}}$

Correct answer:

${\frac{(3^xln(3)^{3}tan(z^{3} - 12z))}{(4y)}}$ $\displaystyle {\frac{(3^xln(3)^{3}tan(z^{3} - 12z))}{(4y)}}$

Explanation:

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

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Example Question #3293 : Calculus 3

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

Possible Answers:

${6xe^{(3z)}sin(y)}$ $\displaystyle {6xe^{(3z)}sin(y)}$

${18x^{3}e^{(6z)}sin(y)^{2}}$ $\displaystyle {18x^{3}e^{(6z)}sin(y)^{2}}$

${6x^{3}e^{(6z)}sin(y)^{2}}$ $\displaystyle {6x^{3}e^{(6z)}sin(y)^{2}}$

${2xe^{(3z)}sin(y) + 3x^{2}e^{(3z)}sin(y)}$ $\displaystyle {2xe^{(3z)}sin(y) + 3x^{2}e^{(3z)}sin(y)}$

Correct answer:

${6xe^{(3z)}sin(y)}$ $\displaystyle {6xe^{(3z)}sin(y)}$

Explanation:

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Example Question #921 : Partial Derivatives

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

Possible Answers:

${-\frac{(144\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{5}}}$ $\displaystyle {-\frac{(144\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{5}}}$

${-\frac{(2239488\cdot 4^{(20x)}sin(z^{4} - 3z)^{5})}{y^{15}}}$ $\displaystyle {-\frac{(2239488\cdot 4^{(20x)}sin(z^{4} - 3z)^{5})}{y^{15}}}$

${-\frac{(7776\cdot 4^{(20x)}sin(z^{4} - 3z)^{5})}{y^{5}}}$ $\displaystyle {-\frac{(7776\cdot 4^{(20x)}sin(z^{4} - 3z)^{5})}{y^{5}}}$

${-\frac{(30\cdot 4^{(4x)}sin(z^{4} - 3z))}{y}}$ $\displaystyle {-\frac{(30\cdot 4^{(4x)}sin(z^{4} - 3z))}{y}}$

Correct answer:

${-\frac{(144\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{5}}}$ $\displaystyle {-\frac{(144\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{5}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=-3\cdot 4^{(4x)}ln(y^{2})sin(z^{4} - 3z)\\&f_{y}=-\frac{(6\cdot 4^{(4x)}sin(z^{4} - 3z))}{y}\\&f_{yy}=\frac{(6\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{2}}\\&f_{yyy}=-\frac{(12\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{3}}\\&f_{yyyy}=\frac{(36\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{4}}\\&f_{yyyyy}=-\frac{(144\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{5}}\end{align*}$ $\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=-3\cdot 4^{(4x)}ln(y^{2})sin(z^{4} - 3z)\\&f_{y}=-\frac{(6\cdot 4^{(4x)}sin(z^{4} - 3z))}{y}\\&f_{yy}=\frac{(6\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{2}}\\&f_{yyy}=-\frac{(12\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{3}}\\&f_{yyyy}=\frac{(36\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{4}}\\&f_{yyyyy}=-\frac{(144\cdot 4^{(4x)}sin(z^{4} - 3z))}{y^{5}}\end{align*}$

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Example Question #3295 : Calculus 3

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

Possible Answers:

${\frac{(ln(4y)^{4}cos(x)^{4}ln(z)^{4}sin(x)^{2})}{(64y^{2}z^{2})}}$ $\displaystyle {\frac{(ln(4y)^{4}cos(x)^{4}ln(z)^{4}sin(x)^{2})}{(64y^{2}z^{2})}}$

${\frac{(cos(x)^{4}ln(z)^{4}sin(x)^{2})}{(64y^{10}z^{3})}}$ $\displaystyle {\frac{(cos(x)^{4}ln(z)^{4}sin(x)^{2})}{(64y^{10}z^{3})}}$

${-\frac{cos(x)}{(2y^{2}z^{2})}}$ $\displaystyle {-\frac{cos(x)}{(2y^{2}z^{2})}}$

${\frac{(ln(4y)cos(x))}{z}- ln(4y)ln(z)sin(x) +\frac{ (cos(x)ln(z))}{y}}$ $\displaystyle {\frac{(ln(4y)cos(x))}{z}- ln(4y)ln(z)sin(x) +\frac{ (cos(x)ln(z))}{y}}$

Correct answer:

${-\frac{cos(x)}{(2y^{2}z^{2})}}$ $\displaystyle {-\frac{cos(x)}{(2y^{2}z^{2})}}$

Explanation:

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

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Example Question #3296 : Calculus 3

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

Possible Answers:

${8e^{(4y)}sin(4x + x^{3})}$ $\displaystyle {8e^{(4y)}sin(4x + x^{3})}$

${64z^{3}e^{(16y)}sin(4x + x^{3})^{4}}$ $\displaystyle {64z^{3}e^{(16y)}sin(4x + x^{3})^{4}}$

${\frac{(z^{6}e^{(16y)}sin(4x + x^{3})^{4})}{4}}$ $\displaystyle {\frac{(z^{6}e^{(16y)}sin(4x + x^{3})^{4})}{4}}$

${ze^{(4y)}sin(4x + x^{3}) + 2z^{2}e^{(4y)}sin(4x + x^{3})}$ $\displaystyle {ze^{(4y)}sin(4x + x^{3}) + 2z^{2}e^{(4y)}sin(4x + x^{3})}$

Correct answer:

${8e^{(4y)}sin(4x + x^{3})}$ $\displaystyle {8e^{(4y)}sin(4x + x^{3})}$

Explanation:

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

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Example Question #3297 : Calculus 3

Find $f_{yyz}$ $\displaystyle f_{yyz}$ of the following function: $\displaystyle x^7z+y^4z^2$

Possible Answers:

$\displaystyle 12y^2z^2$

4y^3z^2 $\displaystyle 4y^3z^2$

24y^2z $\displaystyle 24y^2z$

12y^3z $\displaystyle 12y^3z$

Correct answer:

24y^2z $\displaystyle 24y^2z$

Explanation:

In order to solve, you must take a total of three derivatives: the first is $\frac{\partial }{\partial y}$ $\displaystyle \frac{\partial }{\partial y}$ , then again ${\frac{\partial }{\partial y}}$ $\displaystyle {\frac{\partial }{\partial y}}$ , and finally $\frac{\partial }{\partial z}$ $\displaystyle \frac{\partial }{\partial z}$ ,in that order (the notation in the problem statement dictates that). The first derivative you obtain will be 4y^3z^2 $\displaystyle 4y^3z^2$ (the term with x and z goes away because the derivative with that with respect to y is zero). The subsequent derivative is $\displaystyle 12y^2z^2$ . The final derivative with respect to z is 24y^2z $\displaystyle 24y^2z$ . The rule used for all derivatives is $\frac{\mathrm{d} }{\mathrm{d} x}x^n= nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n= nx^{n-1}$ , and we treat all other variables as constants.

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Example Question #3298 : Calculus 3

Find $f_{xxy}$ $\displaystyle f_{xxy}$ of the following function: $2x^3y+5\sin(x)$ $\displaystyle 2x^3y+5\sin(x)$

Possible Answers:

$12x-5\sin(x)$ $\displaystyle 12x-5\sin(x)$

$12x-\5\sin(y)$ $\displaystyle 12x-\5\sin(y)$

$6xy-5\sin(x)$ $\displaystyle 6xy-5\sin(x)$

$12xy+5\cos(x)$ $\displaystyle 12xy+5\cos(x)$

Correct answer:

$12x-5\sin(x)$ $\displaystyle 12x-5\sin(x)$

Explanation:

From the problem statement we must take three consecutive derivatives $\frac{\partial }{\partial x}, \frac{\partial }{\partial x},\frac{\partial }{\partial y}$ $\displaystyle \frac{\partial }{\partial x}, \frac{\partial }{\partial x},\frac{\partial }{\partial y}$ . The first derivative, treating y like a constant, produces $6x^2y+5\cos(x)$ $\displaystyle 6x^2y+5\cos(x)$ . The derivative of this expression again with respect to x is $12xy-5\sin(x)$ $\displaystyle 12xy-5\sin(x)$ . Finally, the derivative of this expression with respect to y treating x as a constant is $12x-5\sin(x)$ $\displaystyle 12x-5\sin(x)$ .

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Calculus 3 : Partial Derivatives

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