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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 #3271 : Calculus 3

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

Possible Answers:

${-\frac{(28800y^{4}\cdot (\frac{(5y^{4})}{(xz^{2})}-\frac{ (4\cdot 2^{(3z)}xln(y))}{5})\cdot (\frac{(4\cdot 2^{(3z)})}{(5y)}+\frac{ (20y^{3})}{(x^{2}z^{2})})\cdot (\frac{(4\cdot 2^{(3z)})}{(5y^{2})}-\frac{ (60y^{2})}{(x^{2}z^{2})})\cdot (\frac{(4\cdot 2^{(3z)}ln(y))}{5}+\frac{ (5y^{4})}{(x^{2}z^{2})}))}{(x^{6}z^{5})}}$

${-(\frac{(10y^{4}ln(x))}{z^{3}}+\frac{ (6\cdot 2^{(3z)}x^{2}ln(2)ln(y))}{5})\cdot (\frac{(5y^{4})}{(xz^{2})}-\frac{ (4\cdot 2^{(3z)}xln(y))}{5})^{3}\cdot (\frac{(2\cdot 2^{(3z)}x^{2})}{(5y)}-\frac{ (20y^{3}ln(x))}{z^{2}})^{2}}$

${\frac{(240y^{2})}{(x^{3}z^{3})}}$

${\frac{(4\cdot 2^{(3z)}x^{2})}{(5y)}-\frac{ (15y^{4})}{(xz^{2})}+\frac{ (12\cdot 2^{(3z)}xln(y))}{5}-\frac{ (40y^{3}ln(x))}{z^{2}}+\frac{ (10y^{4}ln(x))}{z^{3}}+\frac{ (6\cdot 2^{(3z)}x^{2}ln(2)ln(y))}{5}}$

Correct answer:

${\frac{(240y^{2})}{(x^{3}z^{3})}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot 2^{(3z)}x^{2}ln(y))}{5}-\frac{ (5y^{4}ln(x))}{z^{2}}\\&f_{x}=\frac{(4\cdot 2^{(3z)}xln(y))}{5}-\frac{ (5y^{4})}{(xz^{2})}\\&f_{xx}=\frac{(4\cdot 2^{(3z)}ln(y))}{5}+\frac{ (5y^{4})}{(x^{2}z^{2})}\\&f_{xxy}=\frac{(4\cdot 2^{(3z)})}{(5y)}+\frac{ (20y^{3})}{(x^{2}z^{2})}\\&f_{xxyy}=\frac{(60y^{2})}{(x^{2}z^{2})}-\frac{ (4\cdot 2^{(3z)})}{(5y^{2})}\\&f_{xxyyx}=-\frac{(120y^{2})}{(x^{3}z^{2})}\\&f_{xxyyxz}=\frac{(240y^{2})}{(x^{3}z^{3})}\end{align*}$

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Example Question #3272 : Calculus 3

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

Possible Answers:

${\frac{(4xy^{2}z^{2})}{9}- 2sin(x^{3})e^{(z)}sin(y) +\frac{ (8x^{2}yz^{2})}{9}+ 3x^{2}cos(x^{3})e^{(z)}cos(y)}$

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

${\frac{(8xz^{2})}{9}- 3x^{2}cos(x^{3})e^{(z)}cos(y)}$

${-(\frac{(8xz^{2})}{9}- 3x^{2}cos(x^{3})e^{(z)}cos(y))\cdot (sin(x^{3})e^{(z)}sin(y) -\frac{ (4x^{2}yz^{2})}{9})\cdot (\frac{(4x^{2}z^{2})}{9}- sin(x^{3})e^{(z)}cos(y))}$

Correct answer:

${\frac{(8xz^{2})}{9}- 3x^{2}cos(x^{3})e^{(z)}cos(y)}$

Explanation:

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Example Question #3273 : Calculus 3

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

Possible Answers:

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

${-4\cdot 4^{(3z)}xycos(x^{2})}$

${- 2\cdot 4^{(3z)}ysin(x^{2}) - 2\cdot 4^{(3z)}xy^{2}cos(x^{2})}$

${4\cdot 4^{(6z)}xy^{3}cos(x^{2})sin(x^{2})}$

Correct answer:

${-4\cdot 4^{(3z)}xycos(x^{2})}$

Explanation:

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Example Question #3274 : Calculus 3

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

Possible Answers:

${2y^{3}sin(x^{2})tan(12z + z^{2})\cdot (tan(y^{4})^{2} + 1)\cdot (6y^{2}sin(x^{2})tan(12z + z^{2})\cdot (tan(y^{4})^{2} + 1) + 16y^{6}sin(x^{2})tan(y^{4})tan(12z + z^{2})\cdot (tan(y^{4})^{2} + 1))}$

${4y^{6}sin(x^{2})^{2}tan(12z + z^{2})^{2}\cdot (tan(y^{4})^{2} + 1)^{2}}$

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

${- 6y^{2}sin(x^{2})tan(12z + z^{2})\cdot (tan(y^{4})^{2} + 1) - 16y^{6}sin(x^{2})tan(y^{4})tan(12z + z^{2})\cdot (tan(y^{4})^{2} + 1)}$

Correct answer:

${- 6y^{2}sin(x^{2})tan(12z + z^{2})\cdot (tan(y^{4})^{2} + 1) - 16y^{6}sin(x^{2})tan(y^{4})tan(12z + z^{2})\cdot (tan(y^{4})^{2} + 1)}$

Explanation:

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Example Question #3275 : Calculus 3

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

Possible Answers:

${(4xz^{2}cos(y^{3}) - xy^{2}e^{(x^{2})}e^{(z)})\cdot (y^{2}e^{(x^{2})}e^{(z)} - 4z^{2}cos(y^{3}) + 2x^{2}y^{2}e^{(x^{2})}e^{(z)})\cdot (12y^{2}z^{2}sin(y^{3}) + 2ye^{(x^{2})}e^{(z)} + 4x^{2}ye^{(x^{2})}e^{(z)})}$

${- 12y^{2}z^{2}sin(y^{3}) - 2ye^{(x^{2})}e^{(z)} - 4x^{2}ye^{(x^{2})}e^{(z)}}$

${8xz^{2}cos(y^{3}) - ye^{(x^{2})}e^{(z)} - 6x^{2}y^{2}z^{2}sin(y^{3}) - 2xy^{2}e^{(x^{2})}e^{(z)}}$

${-(4xz^{2}cos(y^{3}) - xy^{2}e^{(x^{2})}e^{(z)})^{2}\cdot (ye^{(x^{2})}e^{(z)} + 6x^{2}y^{2}z^{2}sin(y^{3}))}$

Correct answer:

${- 12y^{2}z^{2}sin(y^{3}) - 2ye^{(x^{2})}e^{(z)} - 4x^{2}ye^{(x^{2})}e^{(z)}}$

Explanation:

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Example Question #3276 : Calculus 3

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

Possible Answers:

${\frac{(64\cdot 3^{(12z)}ln(3y)^{2}e^{(3x)}ln(3)^{3})}{(27y)}}$

${\frac{(4\cdot 3^{(4z)}e^{(x)}ln(3))}{(3y)}}$

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

${\frac{(4\cdot 3^{(12z)}ln(3y)^{2}e^{(3x)}ln(3))}{(27y)}}$

Correct answer:

${\frac{(4\cdot 3^{(4z)}e^{(x)}ln(3))}{(3y)}}$

Explanation:

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Example Question #3277 : Calculus 3

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

Possible Answers:

${\frac{(432z^{2}cos(z^{3})sin(z^{3})cos(y)^{2}\cdot (\frac{(cos(z^{3})cos(y))}{x^{3}}+\frac{ (2ycos(x^{2})sin(z^{2}))}{3})\cdot (\frac{(3cos(z^{3})cos(y))}{x^{4}}+\frac{ (4xysin(x^{2})sin(z^{2}))}{3})\cdot (\frac{(4ysin(x^{2})sin(z^{2}))}{3}-\frac{ (12cos(z^{3})cos(y))}{x^{5}}+\frac{ (8x^{2}ycos(x^{2})sin(z^{2}))}{3})\cdot (\frac{(4sin(x^{2})sin(z^{2}))}{3}+\frac{ (12cos(z^{3})sin(y))}{x^{5}}+\frac{ (8x^{2}cos(x^{2})sin(z^{2}))}{3}))}{x^{10}}}$

${\frac{(3cos(z^{3})cos(y))}{x^{3}}+ 2ycos(x^{2})sin(z^{2}) -\frac{ (6cos(z^{3})sin(y))}{x^{4}}+\frac{ (2y^{2}zcos(x^{2})cos(z^{2}))}{3}-\frac{ (4xy^{2}sin(x^{2})sin(z^{2}))}{3}-\frac{ (3z^{2}sin(z^{3})sin(y))}{x^{3}}}$

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

${\frac{(36z^{2}sin(z^{3})cos(y))}{x^{5}}}$

Correct answer:

${\frac{(36z^{2}sin(z^{3})cos(y))}{x^{5}}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(cos(z^{3})sin(y))}{x^{3}}+\frac{ (y^{2}cos(x^{2})sin(z^{2}))}{3}\\&f_{y}=\frac{(cos(z^{3})cos(y))}{x^{3}}+\frac{ (2ycos(x^{2})sin(z^{2}))}{3}\\&f_{yx}=-\frac{ (3cos(z^{3})cos(y))}{x^{4}}-\frac{ (4xysin(x^{2})sin(z^{2}))}{3}\\&f_{yxx}=\frac{(12cos(z^{3})cos(y))}{x^{5}}-\frac{ (4ysin(x^{2})sin(z^{2}))}{3}-\frac{ (8x^{2}ycos(x^{2})sin(z^{2}))}{3}\\&f_{yxxy}=-\frac{ (4sin(x^{2})sin(z^{2}))}{3}-\frac{ (12cos(z^{3})sin(y))}{x^{5}}-\frac{ (8x^{2}cos(x^{2})sin(z^{2}))}{3}\\&f_{yxxyy}=-\frac{(12cos(z^{3})cos(y))}{x^{5}}\\&f_{yxxyyz}=\frac{(36z^{2}sin(z^{3})cos(y))}{x^{5}}\end{align*}$

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Example Question #3278 : Calculus 3

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

Possible Answers:

${12\cdot 4^{(y^{2})}x^{2}z + 6\cdot 4^{(y^{2})}x^{2}yz^{2}ln(4)}$

${81\cdot 4^{(6y^{2})}x^{12}y^{2}z^{8}ln(4)^{2}}$

{0}

${729\cdot 4^{(6y^{2})}x^{12}y^{2}z^{8}ln(4)^{2}}$

Correct answer:

{0}

Explanation:

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Example Question #3279 : Calculus 3

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

Possible Answers:

${- 2\cdot 4^yln(3z)e^{(x)} -\frac{ (4\cdot 4^ye^{(x)})}{z}- 4\cdot 4^yln(3z)e^{(x)}ln(4)}$

${-\frac{(32\cdot 4^{(5y)}ln(3z)^{2}e^{(5x)}ln(4)^{9})}{z^{5}}}$

${\frac{(2\cdot 4^ye^{(x)}ln(4)^{2})}{z^{2}}}$

${-\frac{(32\cdot 4^{(5y)}ln(3z)^{3}e^{(5x)}ln(4)^{2})}{z^{2}}}$

Correct answer:

${\frac{(2\cdot 4^ye^{(x)}ln(4)^{2})}{z^{2}}}$

Explanation:

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Example Question #3280 : Calculus 3

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

Possible Answers:

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

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

${-\frac{(65536y^{6}tan(y^{4})^{4}\cdot (tan(y^{4})^{2} + 1)^{2})}{(27x^{20}z^{11})}}$

${\frac{(128y^{3}ln(z^{2})^{3}tan(y^{4})^{5}\cdot (tan(y^{4})^{2} + 1))}{(729x^{14}z^{3})}}$

Correct answer:

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

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(ln(z^{2})tan(y^{4}))}{(3x^{2})}\\&f_{z}=\frac{(2tan(y^{4}))}{(3x^{2}z)}\\&f_{zx}=-\frac{(4tan(y^{4}))}{(3x^{3}z)}\\&f_{zxz}=\frac{(4tan(y^{4}))}{(3x^{3}z^{2})}\\&f_{zxzx}=-\frac{(4tan(y^{4}))}{(x^{4}z^{2})}\\&f_{zxzxy}=-\frac{(16y^{3}\cdot (tan(y^{4})^{2} + 1))}{(x^{4}z^{2})}\\&f_{zxzxyz}=\frac{(32y^{3}\cdot (tan(y^{4})^{2} + 1))}{(x^{4}z^{3})}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3271 : Calculus 3

Example Question #3272 : Calculus 3

Example Question #3273 : Calculus 3

Example Question #3274 : Calculus 3

Example Question #3275 : Calculus 3

Example Question #3276 : Calculus 3

Example Question #3277 : Calculus 3

Example Question #3278 : Calculus 3

Example Question #3279 : Calculus 3

Example Question #3280 : Calculus 3

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