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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #643 : Partial Derivatives

Find the partial derivative $\small \small \small f_y$ of $\small \small \small f(x,y)=e^{5x^5+2y^{3/2}}$

Possible Answers:

$\small \small f_y=3\sqrt{y}e^{5x^5+2y^{3/2}}$

$\small \small \small f_y=25x^4e^{5x^5+2y^{3/2}}$

$\small \small \small \small f_y=e^{3\sqrt{y}}$

$\small \small \small f_y=2y^{3/2}e^{5x^5+3\sqrt{y}}$

Correct answer:

$\small \small f_y=3\sqrt{y}e^{5x^5+2y^{3/2}}$

Explanation:

To find the partial derivative $\small \small \small f_y$ of $\small \small \small f(x,y)=e^{5x^5+2y^{3/2}}$ , we need to take its derivative with respect to $\small y$ while holding $\small x$ constant. We use the chain rule to get

$\small \small \small \small \small f_y=\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}e^{5x^5+2y^{3/2}}= e^{5x^5+2y^{3/2}}\cdot \frac{\partial}{\partial y}(5x^5+2y^{3/2})$

$\small \small \small \small \small \small \small =\frac{3}{2}2y^{1/2}e^{5x^5+2y^{3/2}}=3\sqrt{y}e^{5x^5+2y^{3/2}}$

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Example Question #3011 : Calculus 3

Find the partial derivative $\small \small \small f_x$ of $\small f(x,y)=\ln(x^3+2y^2)$

Possible Answers:

$\small \small f_x=\frac{3x^2}{3x^2+4y}$

$\small \small f_x=\ln(3x^2)$

$\small \small f_x=\frac{4y}{x^3+2y^2}$

$\small f_x=\frac{3x^2}{x^3+2y^2}$

Correct answer:

$\small f_x=\frac{3x^2}{x^3+2y^2}$

Explanation:

To find the partial derivative $\small \small \small f_x$ of $\small f(x,y)=\ln(x^3+2y^2)$ , we need to take its derivative with respect to while holding constant. We use that chain rule to get

$\small \small f_x=\frac{\partial}{\partial x} f(x,y)=\frac{\partial}{\partial x}\ln(x^3+2y^2)=\frac{1}{x^3+2y^2}\frac{\partial}{\partial x}(x^3+2y^2)$

$\small =\frac{3x^2}{x^3+2y^2}$

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Example Question #645 : Partial Derivatives

Find the partial derivative $\small \small \small \small f_y$ of $\small f(x,y)=\ln(x^3+2y^2)$

Possible Answers:

$\small \small \small \small f_y=\frac{4y}{3x^2+4y}$

$\small \small \small \small f_y=\frac{3x^2}{x^3+2y^2}$

$\small \small \small \small f_y=\frac{4y}{x^3+2y^2}$

$\small \small \small \small f_y=\ln(4y)$

Correct answer:

$\small \small \small \small f_y=\frac{4y}{x^3+2y^2}$

Explanation:

To find the partial derivative $\small \small \small \small f_y$ of $\small f(x,y)=\ln(x^3+2y^2)$ , we need to take its derivative with respect to $\small y$ while holding $\small x$ constant. We use that chain rule to get

$\small \small \small f_y=\frac{\partial}{\partial y} f(x,y)=\frac{\partial}{\partial y}\ln(x^3+2y^2)=\frac{1}{x^3+2y^2}\frac{\partial}{\partial y}(x^3+2y^2)$

$\small \small =\frac{4y}{x^3+2y^2}$

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Example Question #646 : Partial Derivatives

Find the partial derivative $\small \small \small \small \small f_x$ of $\small \small f(x,y,z)=x^5y^{10}z^{15}$

Possible Answers:

$\small \small \small f_x=5x^4y^{10}z^{15}$

$\small \small \small \small f_x=0$

$\small \small \small \small f_x=5x^5y^{10}z^{15}$

$\small \small \small \small f_x=50x^4y^{9}z^{15}$

Correct answer:

$\small \small \small f_x=5x^4y^{10}z^{15}$

Explanation:

To find the partial derivative $\small \small \small \small \small f_x$ of $\small \small f(x,y,z)=x^5y^{10}z^{15}$ , we need to take its derivative with respect to $\small x$ while holding $\small y,z$ constant. So we have

$\small \small \small f_x=\frac{\partial}{\partial x} f(x,y,z)=\frac{\partial}{\partial x} x^5y^{10}z^{15}=y^{10}z^{15}\frac{\partial}{\partial x}x^5$

$\small =5x^4y^{10}z^{15}$

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Example Question #647 : Partial Derivatives

Find the partial derivative $\small \small \small \small \small \small f_y$ of $\small \small f(x,y,z)=x^5y^{10}z^{15}$

Possible Answers:

$\small \small \small \small f_y=10x^5y^{9}z^{15}$

$\small \small \small \small f_y=x^4y^{9}z^{15}$

$\small \small \small \small f_y=0$

$\small \small \small \small f_y=50x^4y^{9}z^{15}$

Correct answer:

$\small \small \small \small f_y=10x^5y^{9}z^{15}$

Explanation:

To find the partial derivative $\small \small \small \small \small \small f_y$ of $\small \small f(x,y,z)=x^5y^{10}z^{15}$ , we need to take its derivative with respect to $\small y$ while holding $\small \small x,z$ constant. So we have

$\small \small \small \small f_y=\frac{\partial}{\partial y} f(x,y,z)=\frac{\partial}{\partial y} x^5y^{10}z^{15}=x^{5}z^{15}\frac{\partial}{\partial x}y^{10}$

$\small \small \small =10x^5y^{9}z^{15}$

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Example Question #648 : Partial Derivatives

Find the partial derivative $\small \small \small \small \small \small \small f_z$ of $\small \small f(x,y,z)=x^5y^{10}z^{15}$

Possible Answers:

$\small \small \small \small \small \small f_z=15x^5y^{10}z^{13}$

$\small \small \small \small \small f_z=15x^5y^{10}z^{14}$

$\small \small \small \small \small \small f_z=150x^4y^{9}z^{14}$

$\small \small \small \small \small \small f_z=150x^5y^{9}z^{14}$

Correct answer:

$\small \small \small \small \small f_z=15x^5y^{10}z^{14}$

Explanation:

To find the partial derivative $\small \small \small \small \small \small \small f_z$ of $\small \small f(x,y,z)=x^5y^{10}z^{15}$ , we need to take its derivative with respect to $\small z$ while holding $\small \small \small x,y$ constant. So we have

$\small \small \small \small \small f_z=\frac{\partial}{\partial z} f(x,y,z)=\frac{\partial}{\partial z} x^5y^{10}z^{15}=x^{5}y^{10}\frac{\partial}{\partial x}z^{15}$

$\small \small \small \small =15x^5y^{10}z^{14}$

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Example Question #649 : Partial Derivatives

Find the partial derivative $\small \small \small \small \small \small \small \small \small f_y$ of $\small \small \small \small f(x,y,z)=e^{x^2y^{3}z^{10}}$

Possible Answers:

$\small \small \small \small \small \small \small \small f_y=0$

$\small \small \small \small \small \small \small \small f_y=3y^2z^{10}e^{x^2y^3z^{10}}$

$\small \small \small \small \small \small \small \small f_y=3y^2e^{x^2y^3z^{10}}$

$\small \small \small \small \small \small \small f_y=3x^2y^2z^{10}e^{x^2y^3z^{10}}$

Correct answer:

$\small \small \small \small \small \small \small f_y=3x^2y^2z^{10}e^{x^2y^3z^{10}}$

Explanation:

To find the partial derivative $\small \small \small \small \small \small \small \small \small f_y$ of $\small \small \small \small f(x,y,z)=e^{x^2y^{3}z^{10}}$ , we need to take its derivative with respect to $\small y$ while holding $\small \small \small \small \small x,z$ constant. So we use the chain rule to get

$\small \small \small \small \small \small \small f_y=\frac{\partial}{\partial y} f(x,y,z)=\frac{\partial}{\partial y} e^{x^2y^{3}z^{10}}=e^{x^2y^3z^{10}}\frac{\partial}{\partial y}x^2y^3z^{10}$

$\small \small \small \small \small \small =3x^2y^2z^{10}e^{x^2y^3z^{10}}$

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Example Question #650 : Partial Derivatives

Find the partial derivative $\small \small \small \small \small \small \small \small \small \small f_z$ of $\small \small \small \small f(x,y,z)=e^{x^2y^{3}z^{10}}$

Possible Answers:

$\small \small \small \small \small \small \small \small \small f_z=z^{10}e^{x^2y^3z^{10}}$

$\small \small \small \small \small \small \small \small \small f_z=10z^{9}e^{x^2y^3z^{10}}$

$\small \small \small \small \small \small \small \small f_z=10x^2y^3z^{9}e^{x^2y^3z^{10}}$

$\small \small \small \small \small \small \small \small \small f_z=e^{10x^2y^3z^9}$

Correct answer:

$\small \small \small \small \small \small \small \small f_z=10x^2y^3z^{9}e^{x^2y^3z^{10}}$

Explanation:

To find the partial derivative $\small \small \small \small \small \small \small \small \small \small f_z$ of $\small \small \small \small f(x,y,z)=e^{x^2y^{3}z^{10}}$ , we need to take its derivative with respect to $\small z$ while holding $\small \small \small \small \small \small x,y$ constant. So we use the chain rule to get

$\small \small \small \small \small \small \small \small f_z=\frac{\partial}{\partial z} f(x,y,z)=\frac{\partial}{\partial z} e^{x^2y^{3}z^{10}}=e^{x^2y^3z^{10}}\frac{\partial}{\partial z}x^2y^3z^{10}$

$\small \small \small \small \small \small \small =10x^2y^3z^{9}e^{x^2y^3z^{10}}$

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Example Question #3012 : Calculus 3

Find the partial derivative $\small \small \small \small \small \small \small \small f_x$ of $\small \small \small \small f(x,y,z)=e^{x^2y^{3}z^{10}}$

Possible Answers:

$\small \small \small \small \small \small \small f_x=2xy^3e^{x^2y^3z^{10}}$

$\small \small \small \small \small \small \small f_x=e^{2xy^3z^{10}}$

$\small \small \small \small \small \small \small f_x=2xe^{x^2y^3z^{10}}$

$\small \small \small \small \small \small f_x=2xy^3z^{10}e^{x^2y^3z^{10}}$

Correct answer:

$\small \small \small \small \small \small f_x=2xy^3z^{10}e^{x^2y^3z^{10}}$

Explanation:

To find the partial derivative $\small \small \small \small \small \small \small \small f_x$ of $\small \small \small \small f(x,y,z)=e^{x^2y^{3}z^{10}}$ , we need to take its derivative with respect to $\small x$ while holding $\small \small \small \small y,z$ constant. So we use the chain rule to get

$\small \small \small \small \small \small \small f_x=\frac{\partial}{\partial x} f(x,y,z)=\frac{\partial}{\partial x} e^{x^2y^{3}z^{10}}=e^{x^2y^3z^{10}}\frac{\partial}{\partial x}x^2y^3z^{10}$

$\small \small \small \small \small =2xy^3z^{10}e^{x^2y^3z^{10}}$

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Example Question #3013 : Calculus 3

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

Possible Answers:

${-(x^{2}sin(y^{3})sin(z) +\frac{ (2y^{2}ln(4x)sin(4z + z^{3})\cdot (3z^{2} + 4))}{7})\cdot (2xsin(y^{3})cos(z) +\frac{ (2y^{2}cos(4z + z^{3}))}{(7x)})}$

${- 2xsin(y^{3})sin(z) -\frac{ (2y^{2}sin(4z + z^{3})\cdot (3z^{2} + 4))}{(7x)}}$

${2xsin(y^{3})cos(z) - x^{2}sin(y^{3})sin(z) +\frac{ (2y^{2}cos(4z + z^{3}))}{(7x)}-\frac{ (2y^{2}ln(4x)sin(4z + z^{3})\cdot (3z^{2} + 4))}{7}}$

${(x^{2}sin(y^{3})sin(z) +\frac{ (2y^{2}ln(4x)sin(4z + z^{3})\cdot (3z^{2} + 4))}{7})\cdot (2xsin(y^{3})sin(z) +\frac{ (2y^{2}sin(4z + z^{3})\cdot (3z^{2} + 4))}{(7x)})}$

Correct answer:

${- 2xsin(y^{3})sin(z) -\frac{ (2y^{2}sin(4z + z^{3})\cdot (3z^{2} + 4))}{(7x)}}$

Explanation:

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

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #643 : Partial Derivatives

Example Question #3011 : Calculus 3

Example Question #645 : Partial Derivatives

Example Question #646 : Partial Derivatives

Example Question #647 : Partial Derivatives

Example Question #648 : Partial Derivatives

Example Question #649 : Partial Derivatives

Example Question #650 : Partial Derivatives

Example Question #3012 : Calculus 3

Example Question #3013 : Calculus 3

All Calculus 3 Resources

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