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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #3079 : Calculus 3

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

Possible Answers:

${\frac{(cos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (\frac{(6zcos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1))}{5}+\frac{ (2cos(y^{3} - 3y)tan(11x + x^{3})tan(11z + z^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11)^{2})}{5})\cdot (3z^{2} + 11))}{5}}$

${\frac{(2cos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11))}{5}}$

${\frac{(6zcos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1))}{5}+\frac{ (2cos(y^{3} - 3y)tan(11x + x^{3})tan(11z + z^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11)^{2})}{5}}$

${\frac{(cos(y^{3} - 3y)^{2}tan(11x + x^{3})^{2}\cdot (tan(11z + z^{3})^{2} + 1)^{2}\cdot (3z^{2} + 11)^{2})}{25}}$

Correct answer:

${\frac{(6zcos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1))}{5}+\frac{ (2cos(y^{3} - 3y)tan(11x + x^{3})tan(11z + z^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11)^{2})}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(cos(y^{3} - 3y)tan(11x + x^{3})tan(11z + z^{3}))}{5}\\&f_{z}=\frac{(cos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11))}{5}\\&f_{zz}=\frac{(6zcos(y^{3} - 3y)tan(11x + x^{3})\cdot (tan(11z + z^{3})^{2} + 1))}{5}+\frac{ (2cos(y^{3} - 3y)tan(11x + x^{3})tan(11z + z^{3})\cdot (tan(11z + z^{3})^{2} + 1)\cdot (3z^{2} + 11)^{2})}{5}\end{align*}$

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Example Question #712 : Partial Derivatives

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

Possible Answers:

${-(2\cdot 4^{(3x)}cos(4z + z^{4})cos(y) - 16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x))\cdot (2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x))}$

${32y^{3}e^{(y^{4})}e^{(z^{4})}sin(x) - 4\cdot 4^{(3x)}cos(4z + z^{4})cos(y)}$

${2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)}$

${(2\cdot 4^{(3x)}cos(4z + z^{4})cos(y) - 16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x))^{2}}$

Correct answer:

${2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=4e^{(y^{4})}e^{(z^{4})}sin(x) - 2\cdot 4^{(3x)}cos(4z + z^{4})sin(y)\\&f_{y}=16y^{3}e^{(y^{4})}e^{(z^{4})}sin(x) - 2\cdot 4^{(3x)}cos(4z + z^{4})cos(y)\\&f_{yy}=2\cdot 4^{(3x)}cos(4z + z^{4})sin(y) + 48y^{2}e^{(y^{4})}e^{(z^{4})}sin(x) + 64y^{6}e^{(y^{4})}e^{(z^{4})}sin(x)\end{align*}$

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Example Question #3081 : Calculus 3

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

Possible Answers:

${(\frac{(6xe^{(y^{2})})}{z}+\frac{ (e^{(4y)}cos(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5})\cdot (3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5})}$

${(3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5})\cdot (\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5})}$

${\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}}$

${3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5}+\frac{ (6xe^{(y^{2})})}{z}+\frac{ (e^{(4y)}cos(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}}$

Correct answer:

${\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=3xln(z^{2})e^{(y^{2})} +\frac{ (e^{(4y)}tan(z^{2} - 11z)cos(x))}{5}\\&f_{x}=3ln(z^{2})e^{(y^{2})} -\frac{ (e^{(4y)}tan(z^{2} - 11z)sin(x))}{5}\\&f_{xz}=\frac{(6e^{(y^{2})})}{z}-\frac{ (e^{(4y)}sin(x)\cdot (2z - 11)\cdot (tan(z^{2} - 11z)^{2} + 1))}{5}\end{align*}$

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Example Question #3082 : Calculus 3

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

Possible Answers:

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

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

${(\frac{(4\cdot 4^{(3x)}ye^{(z)})}{3}+ 10x^{4}ycos(y^{2})cos(z^{3}))^{2}}$

${\frac{(8\cdot 4^{(3x)}ye^{(z)})}{3}+ 20x^{4}ycos(y^{2})cos(z^{3})}$

Correct answer:

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

Explanation:

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

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Example Question #3083 : Calculus 3

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

Possible Answers:

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

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

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

${\frac{(4\cdot 3^{(4z)}\cdot 4^x)}{(9y)}-\frac{ (6z^{3}e^{(y)})}{x}}$

Correct answer:

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

Explanation:

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

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Example Question #3084 : Calculus 3

Evaluate $\frac{\partial}{\partial x}(x^2+z\ln(xy))$

Possible Answers:

2x+z

$2x+\frac{z}{xy}$

$2x+\frac{z}{x}$

$\frac{z}{xy}$

Correct answer:

$2x+\frac{z}{x}$

Explanation:

We proceed as follows, using the rules of partial differentiation.

$\frac{\partial}{\partial x}(x^2+z\ln(xy))$ . Start

$=\frac{\partial}{\partial x}(x^2)+\frac{\partial}{\partial x}z\ln(xy)$ . Sum Rule

$=2x+z\frac{1}{xy}y$ . Chain Rule and Power Rule, treating y,z as constants.

$=2x+\frac{z}{y}$ . Simplify

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Example Question #3085 : Calculus 3

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

Possible Answers:

${\frac{(12y^{2}z^{2})}{x^{2}}-\frac{ (4\cdot 4^{(3z)}cos(y^{2} - 4y))}{x}-\frac{ (8y^{2}z^{3})}{x^{3}}- 6\cdot 4^{(3z)}ln(x^{2})ln(4)cos(y^{2} - 4y)}$

${(\frac{(4\cdot 4^{(3z)}cos(y^{2} - 4y))}{x}+\frac{ (8y^{2}z^{3})}{x^{3}})\cdot (\frac{(24y^{2}z^{2})}{x^{3}}+\frac{ (12\cdot 4^{(3z)}ln(4)cos(y^{2} - 4y))}{x})}$

${-(\frac{(12y^{2}z^{2})}{x^{2}}- 6\cdot 4^{(3z)}ln(x^{2})ln(4)cos(y^{2} - 4y))\cdot (\frac{(4\cdot 4^{(3z)}cos(y^{2} - 4y))}{x}+\frac{ (8y^{2}z^{3})}{x^{3}})}$

${-\frac{ (24y^{2}z^{2})}{x^{3}}-\frac{ (12\cdot 4^{(3z)}ln(4)cos(y^{2} - 4y))}{x}}$

Correct answer:

${-\frac{ (24y^{2}z^{2})}{x^{3}}-\frac{ (12\cdot 4^{(3z)}ln(4)cos(y^{2} - 4y))}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(4y^{2}z^{3})}{x^{2}}- 2\cdot 4^{(3z)}ln(x^{2})cos(y^{2} - 4y)\\&f_{x}=-\frac{ (4\cdot 4^{(3z)}cos(y^{2} - 4y))}{x}-\frac{ (8y^{2}z^{3})}{x^{3}}\\&f_{xz}=-\frac{ (24y^{2}z^{2})}{x^{3}}-\frac{ (12\cdot 4^{(3z)}ln(4)cos(y^{2} - 4y))}{x}\end{align*}$

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Example Question #3086 : Calculus 3

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

Possible Answers:

${-\frac{(32x^{4}ln(z^{2})cos(4y + y^{4})sin(4y + y^{4})\cdot (4y^{3} + 4))}{z}}$

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

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

${\frac{(32x^{4}ln(z^{2})sin(4y + y^{4})^{2}\cdot (4y^{3} + 4)^{2})}{z}}$

Correct answer:

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

Explanation:

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

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Example Question #3087 : Calculus 3

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

Possible Answers:

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

${-\frac{(4ln(4z)cos(x^{2} - 3x)cos(3y + y^{2})^{2}sin(x^{2} - 3x)\cdot (2x - 3))}{(81z)}}$

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

${\frac{(4ln(4z)cos(3y + y^{2})^{2}sin(x^{2} - 3x)^{2}\cdot (2x - 3)^{2})}{(81z)}}$

Correct answer:

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

Explanation:

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

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Example Question #3088 : Calculus 3

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

Possible Answers:

${(\frac{(ysin(3z + z^{4}))}{x}+ 16y^{3}cos(z^{2})sin(x^{3})e^{(y^{4})})\cdot (\frac{sin(3z + z^{4})}{x}+ 48y^{2}cos(z^{2})sin(x^{3})e^{(y^{4})} + 64y^{6}cos(z^{2})sin(x^{3})e^{(y^{4})})}$

${\frac{sin(3z + z^{4})}{x}+ 48y^{2}cos(z^{2})sin(x^{3})e^{(y^{4})} + 64y^{6}cos(z^{2})sin(x^{3})e^{(y^{4})}}$

${(\frac{(ysin(3z + z^{4}))}{x}+ 16y^{3}cos(z^{2})sin(x^{3})e^{(y^{4})})^{2}}$

${\frac{(2ysin(3z + z^{4}))}{x}+ 32y^{3}cos(z^{2})sin(x^{3})e^{(y^{4})}}$

Correct answer:

${\frac{sin(3z + z^{4})}{x}+ 48y^{2}cos(z^{2})sin(x^{3})e^{(y^{4})} + 64y^{6}cos(z^{2})sin(x^{3})e^{(y^{4})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=4cos(z^{2})sin(x^{3})e^{(y^{4})} +\frac{ (y^{2}sin(3z + z^{4}))}{(2x)}\\&f_{y}=\frac{(ysin(3z + z^{4}))}{x}+ 16y^{3}cos(z^{2})sin(x^{3})e^{(y^{4})}\\&f_{yy}=\frac{sin(3z + z^{4})}{x}+ 48y^{2}cos(z^{2})sin(x^{3})e^{(y^{4})} + 64y^{6}cos(z^{2})sin(x^{3})e^{(y^{4})}\end{align*}$

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Calculus 3 : Partial Derivatives

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Example Question #3079 : Calculus 3

Example Question #712 : Partial Derivatives

Example Question #3081 : Calculus 3

Example Question #3082 : Calculus 3

Example Question #3083 : Calculus 3

Example Question #3084 : Calculus 3

Example Question #3085 : Calculus 3

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