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Calculus 3 : Calculus 3

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Example Questions

Example Question #3051 : Calculus 3

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

Possible Answers:

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

${-16\cdot 4^{(y^{2})}ye^{(4x)}ln(4)cos(z)\cdot (\frac{(4e^{(z)})}{y}- 4\cdot 4^{(y^{2})}ye^{(4x)}ln(4)cos(z))}$

${-16\cdot 4^{(y^{2})}ye^{(4x)}ln(4)cos(z)}$

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

Correct answer:

${-16\cdot 4^{(y^{2})}ye^{(4x)}ln(4)cos(z)}$

Explanation:

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

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Example Question #3052 : Calculus 3

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\\&\text{Where }f(x,y,z)=\frac{(2sin(x^{2})e^{(z^{2})}tan(11y + y^{3}))}{7}-\frac{ (z^{4}ln(x^{2}))}{y^{3}}\end{align*}$

Possible Answers:

${\frac{(6z^{4})}{(xy^{4})}+\frac{ (4xcos(x^{2})e^{(z^{2})}\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11))}{7}}$

${-(\frac{(2z^{4})}{(xy^{3})}-\frac{ (4xcos(x^{2})e^{(z^{2})}tan(11y + y^{3}))}{7})\cdot (\frac{(3z^{4}ln(x^{2}))}{y^{4}}+\frac{ (2sin(x^{2})e^{(z^{2})}\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11))}{7})}$

${-(\frac{(2z^{4})}{(xy^{3})}-\frac{ (4xcos(x^{2})e^{(z^{2})}tan(11y + y^{3}))}{7})\cdot (\frac{(6z^{4})}{(xy^{4})}+\frac{ (4xcos(x^{2})e^{(z^{2})}\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11))}{7})}$

${\frac{(3z^{4}ln(x^{2}))}{y^{4}}-\frac{ (2z^{4})}{(xy^{3})}+\frac{ (2sin(x^{2})e^{(z^{2})}\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11))}{7}+\frac{ (4xcos(x^{2})e^{(z^{2})}tan(11y + y^{3}))}{7}}$

Correct answer:

${\frac{(6z^{4})}{(xy^{4})}+\frac{ (4xcos(x^{2})e^{(z^{2})}\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11))}{7}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2sin(x^{2})e^{(z^{2})}tan(11y + y^{3}))}{7}-\frac{ (z^{4}ln(x^{2}))}{y^{3}}\\&f_{x}=\frac{(4xcos(x^{2})e^{(z^{2})}tan(11y + y^{3}))}{7}-\frac{ (2z^{4})}{(xy^{3})}\\&f_{xy}=\frac{(6z^{4})}{(xy^{4})}+\frac{ (4xcos(x^{2})e^{(z^{2})}\cdot (tan(11y + y^{3})^{2} + 1)\cdot (3y^{2} + 11))}{7}\end{align*}$

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Example Question #3053 : Calculus 3

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

Possible Answers:

${-\frac{(32e^{(6z)}\cdot (2x - 12)^{2}\cdot (tan(x^{2} - 12x)^{2} + 1)^{2})}{y^{5}}}$

${\frac{(4e^{(3z)}\cdot (20y^{3}e^{(y^{4})}e^{(z^{3})} -\frac{ (8e^{(3z)}tan(x^{2} - 12x))}{y^{3}})\cdot (2x - 12)\cdot (tan(x^{2} - 12x)^{2} + 1))}{y^{2}}}$

${20y^{3}e^{(y^{4})}e^{(z^{3})} -\frac{ (8e^{(3z)}tan(x^{2} - 12x))}{y^{3}}+\frac{ (4e^{(3z)}\cdot (2x - 12)\cdot (tan(x^{2} - 12x)^{2} + 1))}{y^{2}}}$

${-\frac{(8e^{(3z)}\cdot (2x - 12)\cdot (tan(x^{2} - 12x)^{2} + 1))}{y^{3}}}$

Correct answer:

${-\frac{(8e^{(3z)}\cdot (2x - 12)\cdot (tan(x^{2} - 12x)^{2} + 1))}{y^{3}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=5e^{(y^{4})}e^{(z^{3})} +\frac{ (4e^{(3z)}tan(x^{2} - 12x))}{y^{2}}\\&f_{x}=\frac{(4e^{(3z)}\cdot (2x - 12)\cdot (tan(x^{2} - 12x)^{2} + 1))}{y^{2}}\\&f_{xy}=-\frac{(8e^{(3z)}\cdot (2x - 12)\cdot (tan(x^{2} - 12x)^{2} + 1))}{y^{3}}\end{align*}$

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Example Question #3054 : Calculus 3

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

Possible Answers:

${-9\cdot 4^xz^{2}\cdot (3x^{4}sin(z) - 9\cdot 4^xyz^{2})}$

${18\cdot 4^xyz^{2} - 6x^{4}sin(z)}$

${(3x^{4}sin(z) - 9\cdot 4^xyz^{2})^{2}}$

${9\cdot 4^xz^{2}}$

Correct answer:

${9\cdot 4^xz^{2}}$

Explanation:

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Example Question #3055 : Calculus 3

Find the partial derivative

$\frac{\partial f}{\partial x}$

of the function

$f(x,y)=e^{2x+3y}$

Possible Answers:

$\frac{\partial f}{\partial x}=5e^{2x+3y}$

$\frac{\partial f}{\partial x}=2e^{2x+3y}$

$\frac{\partial f}{\partial x}=3e^{2x+3y}$

$\frac{\partial f}{\partial x}=6e^{2x+3y}$

Correct answer:

$\frac{\partial f}{\partial x}=2e^{2x+3y}$

Explanation:

To find the partial derivative with respect to , or

$\frac{\partial f }{\partial x}$

we take the derivative with respect to and treat the other variables as constants.

$\frac{\partial f }{\partial x}=\frac{\partial }{\partial x}(e^{2x+3y})=2e^{2x+3y}$

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Example Question #3056 : Calculus 3

Find the partial derivative

$\frac{\partial f}{\partial y}$

of the function

f(x,y)=xcos(y)

Possible Answers:

$\frac{\partial f}{\partial y}=\frac{1}{2}x^2cos(y)$

$\frac{\partial f}{\partial y}=-xsin(y)$

$\frac{\partial f}{\partial y}=xsin(y)$

$\frac{\partial f}{\partial y}=cos(y)$

Correct answer:

$\frac{\partial f}{\partial y}=-xsin(y)$

Explanation:

To find the partial derivative with respect to , or

$\frac{\partial f }{\partial y}$

we take the derivative with respect to and treat the other variables as constants.

$\frac{\partial f }{\partial y}=\frac{\partial }{\partial y}[xcos(y)]=-xsin(y)$

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Example Question #3057 : Calculus 3

Find the partial derivative with respect to , $\frac{\partial F}{\partial x}$ , of the function

F(x,y)=cos( x) + sin (y) +2xy

Possible Answers:

$\frac{\partial F}{\partial x}=-sin(x)-cos(y)$

$\frac{\partial F}{\partial x}=-sin(x)+2y$

$\frac{\partial F}{\partial x}=cos(y)+2x$

$\frac{\partial F}{\partial x}=cos(x)+cos(y)+2$

Correct answer:

$\frac{\partial F}{\partial x}=-sin(x)+2y$

Explanation:

To find The partial derivative with respect to

$\frac{\partial F}{\partial x}$

you derive the function with respect to and treat the other variables as constants.

$\frac{\partial F}{\partial x}=\frac{\partial }{\partial x}[cos( x) + sin (y) +2xy]=-sin(x)+2y$

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Example Question #3058 : Calculus 3

Find the partial derivative with respect to , $\frac{\partial F}{\partial y}$ , of the function

F(x,y)=tan (x) - cot (y) + e^y

Possible Answers:

$\frac{\partial F}{\partial y}=csc^2(y)+e^y$

$\frac{\partial F}{\partial y}=sec^2(x)$

$\frac{\partial F}{\partial y}=e^y$

$\frac{\partial F}{\partial y}=sex^2(x)+csc^2(y)+e^y$

Correct answer:

$\frac{\partial F}{\partial y}=csc^2(y)+e^y$

Explanation:

To find The partial derivative with respect to

$\frac{\partial F}{\partial y}$

you derive the function with respect to and treat the other variables as constants.

$\frac{\partial F}{\partial y}=\frac{\partial }{\partial y}[tan (x) - cot (y) + e^y]=csc^2(y)+e^y$

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Example Question #3059 : Calculus 3

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

Possible Answers:

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

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

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

-sec^2(4x)sin(y^2)cos(z^3-3z)(3z^2-3)

Correct answer:

-sec^2(4x)sin(y^2)cos(z^3-3z)(3z^2-3)

Explanation:

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

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Example Question #3060 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3051 : Calculus 3

Example Question #3052 : Calculus 3

Example Question #3053 : Calculus 3

Example Question #3054 : Calculus 3

Example Question #3055 : Calculus 3

Example Question #3056 : Calculus 3

Example Question #3057 : Calculus 3

Example Question #3058 : Calculus 3

Example Question #3059 : Calculus 3

Example Question #3060 : Calculus 3

All Calculus 3 Resources

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