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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 #923 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #924 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #925 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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Example Question #926 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #927 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #928 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #929 : Partial Derivatives

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

Possible Answers:

12y^2z^2

4y^3z^2

12y^3z

24y^2z

Correct answer:

24y^2z

Explanation:

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

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Example Question #930 : Partial Derivatives

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

Possible Answers:

$12x-5\sin(x)$

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

$6xy-5\sin(x)$

$12xy+5\cos(x)$

Correct answer:

$12x-5\sin(x)$

Explanation:

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

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Example Question #931 : Partial Derivatives

Find $f_{yx}$ of the following function:

$f(x, y, z)=xy^2z+x\sec(y)$

Possible Answers:

$2xyz+x\sec(y)\tan(y)$

$2yz-\sec(y)\tan(y)$

$2yz+\sec(y)\tan(y)$

Correct answer:

$2yz+\sec(y)\tan(y)$

Explanation:

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

To start, we must take the derivative of the function with respect to y:

$f_y=2xyz+x\sec(y)\tan(y)$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Finally, we take the derivative of the above function with respect to x:

$f_{yx}=2yz+\sec(y)\tan(y)$

The derivative was found using rules above as well as

$\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$

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Example Question #932 : Partial Derivatives

Find $f_{yx}$ for the following function:

$f(x, y, z)=ze^{xy}$

Possible Answers:

$zx^2e^{xy}$

$zxe^{xy}$

$ze^{xy}+zx^2e^{xy}$

$zxe^{xy}+zx^2e^{xy}$

Correct answer:

$ze^{xy}+zx^2e^{xy}$

Explanation:

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

To start, we must find the derivative of the function with respect to x:

$f_y=zxe^{xy}$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Finally, we find the derivative of the above function with respect to x:

$f_{yx}=ze^{xy}+zx^2e^{xy}$

We used the above rules to find the derivative as well as the following:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #923 : Partial Derivatives

Example Question #924 : Partial Derivatives

Example Question #925 : Partial Derivatives

Example Question #926 : Partial Derivatives

Example Question #927 : Partial Derivatives

Example Question #928 : Partial Derivatives

Example Question #929 : Partial Derivatives

Example Question #930 : Partial Derivatives

Example Question #931 : Partial Derivatives

Example Question #932 : Partial Derivatives

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