Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #2871 : Calculus 3

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

Possible Answers:

\(\displaystyle {-8xzsin{(z^2)}e^{(x^2)}ln{(y)}}\)

\(\displaystyle {32xz^2sin{(z^2)}^2e^{(2x^2)}ln{(y)}^2}\)

\(\displaystyle {4xcos{(z^2)}e^{(x^2)}ln{(y)} - 4zsin{(z^2)}e^{(x^2)}ln{(y)}}\)

\(\displaystyle -16xzcos{(z^2)}sin{(z^2)}e^{(2x^2)}ln{(y)}^2\)

Correct answer:

\(\displaystyle {-8xzsin{(z^2)}e^{(x^2)}ln{(y)}}\)

Explanation:

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

Example Question #2872 : Calculus 3

\(\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{yxy}\\&\text{Where }f(x,y,z)=\frac{{(4\cdot4^{(z^4)}ln{(y^4)})}}{x^2}\end{align*}\)

Possible Answers:

\(\displaystyle \frac{-{(2048\cdot4^{(3z^4)}ln{(y^4)})}}{{(x^7y^2)}}\)

\(\displaystyle {\frac{{(32\cdot4^{(z^4)})}}{{(x^3y^2)}}}\)

\(\displaystyle {\frac{{(32\cdot4^{(z^4)})}}{{(x^2y)}} -\frac{ {(8\cdot4^{(z^4)}ln{(y^4)})}}{x^3}}\)

\(\displaystyle {\frac{-{(16384\cdot4^{(3z^4)})}}{{(x^8y^4)}}}\)

Correct answer:

\(\displaystyle {\frac{{(32\cdot4^{(z^4)})}}{{(x^3y^2)}}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f_{y}=\frac{{(16\cdot4^{(z^4)})}}{{(x^2y)}}\\&f_{yx}=\frac{-{(32\cdot4^{(z^4)})}}{{(x^3y)}}\\&f_{yxy}=\frac{{(32\cdot4^{(z^4)})}}{{(x^3y^2)}}\end{align*}\)

Example Question #2873 : Calculus 3

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

Possible Answers:

\(\displaystyle \frac{-{(1152y^9z^4cos{(z^3)}^2ln{(x^2)}^2sin{(z^3)})}}{x}\)

\(\displaystyle \frac{{(1152y^9z^4cos{(z^3)}^2ln{(x^2)}^2sin{(z^3)})}}{x}\)

\(\displaystyle {-\frac{ {(8y^3sin{(z^3)})}}{x} - 24y^3z^2cos{(z^3)}ln{(x^2)}}\)

\(\displaystyle {\frac{{(72y^3z^4sin{(z^3)})}}{x} -\frac{ {(48y^3zcos{(z^3)})}}{x}}\)

Correct answer:

\(\displaystyle {\frac{{(72y^3z^4sin{(z^3)})}}{x} -\frac{ {(48y^3zcos{(z^3)})}}{x}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f_{x}=\frac{-{(8y^3sin{(z^3)})}}{x}\\&f_{xz}=\frac{-{(24y^3z^2cos{(z^3)})}}{x}\\&f_{xzz}=\frac{{(72y^3z^4sin{(z^3)})}}{x} -\frac{ {(48y^3zcos{(z^3)})}}{x}\end{align*}\)

Example Question #2874 : Calculus 3

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

Possible Answers:

\(\displaystyle {\frac{-{(8zcos{(z^2)}ln{(x^2)}^2sin{(z^2)})}}{y^4}}\)

\(\displaystyle {\frac{{(2cos{(z^2)}ln{(x^2)})}}{y^2} +\frac{ {(4zln{(x^2)}sin{(z^2)})}}{y}}\)

\(\displaystyle \frac{{(8zcos{(z^2)}ln{(x^2)}^2sin{(z^2)})}}{y^3}\)

\(\displaystyle {\frac{-{(4zln{(x^2)}sin{(z^2)})}}{y^2}}\)

Correct answer:

\(\displaystyle {\frac{-{(4zln{(x^2)}sin{(z^2)})}}{y^2}}\)

Explanation:

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

Example Question #2875 : Calculus 3

\(\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{zy}\\&\text{Where }f(x,y,z)=-7\cdot3^{(y^4)}sin{(z^4)}e^{(x^4)}\end{align*}\)

Possible Answers:

\(\displaystyle {- 28\cdot3^{(y^4)}z^3cos{(z^4)}e^{(x^4)} - 28\cdot3^{(y^4)}y^3sin{(z^4)}e^{(x^4)}ln{(3)}}\)

\(\displaystyle 784\cdot3^{(2y^4)}y^3z^3cos{(z^4)}sin{(z^4)}e^{(2x^4)}ln{(3)}\)

\(\displaystyle {-112\cdot3^{(y^4)}y^3z^3cos{(z^4)}e^{(x^4)}ln{(3)}}\)

\(\displaystyle {3136\cdot3^{(2y^4)}y^3z^6cos{(z^4)}^2e^{(2x^4)}ln{(3)}}\)

Correct answer:

\(\displaystyle {-112\cdot3^{(y^4)}y^3z^3cos{(z^4)}e^{(x^4)}ln{(3)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f_{z}=-28\cdot3^{(y^4)}z^3cos{(z^4)}e^{(x^4)}\\&f_{zy}=-112\cdot3^{(y^4)}y^3z^3cos{(z^4)}e^{(x^4)}ln{(3)}\end{align*}\)

Example Question #2876 : Calculus 3

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

Possible Answers:

\(\displaystyle {54z^2sin{(x^2)}e^{(y^2)} + 18yz^3sin{(x^2)}e^{(y^2)}}\)

\(\displaystyle {104976y^3z^6sin{(x^2)}^3e^{(3y^2)}}\)

\(\displaystyle 13122yz^7sin{(x^2)}^3e^{(3y^2)}\)

\(\displaystyle {108yzsin{(x^2)}e^{(y^2)}}\)

Correct answer:

\(\displaystyle {108yzsin{(x^2)}e^{(y^2)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f_{y}=18yz^3sin{(x^2)}e^{(y^2)}\\&f_{yz}=54yz^2sin{(x^2)}e^{(y^2)}\\&f_{yzz}=108yzsin{(x^2)}e^{(y^2)}\end{align*}\)

Example Question #509 : Partial Derivatives

\(\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{zxzz}\\&\text{Where }f(x,y,z)=10\cdot2^{(y^2)}cos{(x^4)}cos{(z^4)}\end{align*}\)

Possible Answers:

\(\displaystyle {- 40\cdot2^{(y^2)}x^3cos{(z^4)}sin{(x^4)} - 120\cdot2^{(y^2)}z^3cos{(x^4)}sin{(z^4)}}\)

\(\displaystyle {960\cdot2^{(y^2)}x^3zsin{(x^4)}sin{(z^4)} + 5760\cdot2^{(y^2)}x^3z^5cos{(z^4)}sin{(x^4)} - 2560\cdot2^{(y^2)}x^3z^9sin{(x^4)}sin{(z^4)}}\)

\(\displaystyle {-6400\cdot2^{(2y^2)}x^3z^6cos{(x^4)}sin{(x^4)}sin{(z^4)}^2\cdot{(640\cdot2^{(y^2)}x^3z^6cos{(z^4)}sin{(x^4)} + 480\cdot2^{(y^2)}x^3z^2sin{(x^4)}sin{(z^4)})}\cdot{(960\cdot2^{(y^2)}x^3zsin{(x^4)}sin{(z^4)} + 5760\cdot2^{(y^2)}x^3z^5cos{(z^4)}sin{(x^4)} - 2560\cdot2^{(y^2)}x^3z^9sin{(x^4)}sin{(z^4)})}}\)

\(\displaystyle 2560000\cdot2^{(4y^2)}x^3z^9cos{(x^4)}^3cos{(z^4)}sin{(x^4)}sin{(z^4)}^3\)

Correct answer:

\(\displaystyle {960\cdot2^{(y^2)}x^3zsin{(x^4)}sin{(z^4)} + 5760\cdot2^{(y^2)}x^3z^5cos{(z^4)}sin{(x^4)} - 2560\cdot2^{(y^2)}x^3z^9sin{(x^4)}sin{(z^4)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f_{z}=-40\cdot2^{(y^2)}z^3cos{(x^4)}sin{(z^4)}\\&f_{zx}=160\cdot2^{(y^2)}x^3z^3sin{(x^4)}sin{(z^4)}\\&f_{zxz}=640\cdot2^{(y^2)}x^3z^6cos{(z^4)}sin{(x^4)} + 480\cdot2^{(y^2)}x^3z^2sin{(x^4)}sin{(z^4)}\\&f_{zxzz}=960\cdot2^{(y^2)}x^3zsin{(x^4)}sin{(z^4)} + 5760\cdot2^{(y^2)}x^3z^5cos{(z^4)}sin{(x^4)} - 2560\cdot2^{(y^2)}x^3z^9sin{(x^4)}sin{(z^4)}\end{align*}\)

Example Question #2877 : Calculus 3

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

Possible Answers:

\(\displaystyle {\frac{-{(24y^3cos{(z^4)}sin{(y^4)})}}{x}}\)

\(\displaystyle {\frac{{(6cos{(y^4)}cos{(z^4)})}}{x} - 8y^3cos{(z^4)}ln{(x^3)}sin{(y^4)}}\)

\(\displaystyle \frac{-{(48y^3cos{(y^4)}cos{(z^4)}^2ln{(x^3)}sin{(y^4)})}}{x}\)

\(\displaystyle {\frac{{(192y^6cos{(z^4)}^2ln{(x^3)}sin{(y^4)}^2)}}{x}}\)

Correct answer:

\(\displaystyle {\frac{-{(24y^3cos{(z^4)}sin{(y^4)})}}{x}}\)

Explanation:

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

Example Question #2871 : Calculus 3

\(\displaystyle \begin{align*}&\text{Perform the partial derivation, }f_{zzyx}\\&\text{Where }f(x,y,z)=-7cos{(y^3)}sin{(z^2)}sin{(x)}\end{align*}\)

Possible Answers:

\(\displaystyle {42y^2cos{(z^2)}sin{(y^3)}cos{(x)} - 84y^2z^2sin{(y^3)}sin{(z^2)}cos{(x)}}\)

\(\displaystyle {14zcos{(y^3)}cos{(z^2)}sin{(x)}\cdot{(42y^2cos{(z^2)}sin{(y^3)}cos{(x)} - 84y^2z^2sin{(y^3)}sin{(z^2)}cos{(x)})}\cdot{(42y^2cos{(z^2)}sin{(y^3)}sin{(x)} - 84y^2z^2sin{(y^3)}sin{(z^2)}sin{(x)})}\cdot{(14cos{(y^3)}cos{(z^2)}sin{(x)} - 28z^2cos{(y^3)}sin{(z^2)}sin{(x)})}}\)

\(\displaystyle -28812y^2z^2cos{(y^3)}^3cos{(z^2)}^2sin{(y^3)}sin{(z^2)}^2cos{(x)}sin{(x)}^3\)

\(\displaystyle {21y^2sin{(y^3)}sin{(z^2)}sin{(x)} - 28zcos{(y^3)}cos{(z^2)}sin{(x)} - 7cos{(y^3)}sin{(z^2)}cos{(x)}}\)

Correct answer:

\(\displaystyle {42y^2cos{(z^2)}sin{(y^3)}cos{(x)} - 84y^2z^2sin{(y^3)}sin{(z^2)}cos{(x)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f_{z}=-14zcos{(y^3)}cos{(z^2)}sin{(x)}\\&f_{zz}=28z^2cos{(y^3)}sin{(z^2)}sin{(x)} - 14cos{(y^3)}cos{(z^2)}sin{(x)}\\&f_{zzy}=42y^2cos{(z^2)}sin{(y^3)}sin{(x)} - 84y^2z^2sin{(y^3)}sin{(z^2)}sin{(x)}\\&f_{zzyx}=42y^2cos{(z^2)}sin{(y^3)}cos{(x)} - 84y^2z^2sin{(y^3)}sin{(z^2)}cos{(x)}\end{align*}\)

Example Question #2878 : Calculus 3

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

Possible Answers:

\(\displaystyle {12y^2z^2sin{(x^3)}e^{(y^3)}\cdot{(24yz^2sin{(x^3)}e^{(y^3)} + 36y^4z^2sin{(x^3)}e^{(y^3)})}\cdot{(72x^2yz^2cos{(x^3)}e^{(y^3)} + 108x^2y^4z^2cos{(x^3)}e^{(y^3)})}}\)

\(\displaystyle 1728x^2y^4z^6cos{(x^3)}sin{(x^3)}^2e^{(3y^3)}\)

\(\displaystyle {72x^2yz^2cos{(x^3)}e^{(y^3)} + 108x^2y^4z^2cos{(x^3)}e^{(y^3)}}\)

\(\displaystyle {12x^2z^2cos{(x^3)}e^{(y^3)} + 24y^2z^2sin{(x^3)}e^{(y^3)}}\)

Correct answer:

\(\displaystyle {72x^2yz^2cos{(x^3)}e^{(y^3)} + 108x^2y^4z^2cos{(x^3)}e^{(y^3)}}\)

Explanation:

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f_{y}=12y^2z^2sin{(x^3)}e^{(y^3)}\\&f_{yy}=24yz^2sin{(x^3)}e^{(y^3)} + 36y^4z^2sin{(x^3)}e^{(y^3)}\\&f_{yyx}=72x^2yz^2cos{(x^3)}e^{(y^3)} + 108x^2y^4z^2cos{(x^3)}e^{(y^3)}\end{align*}\)

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