\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2x^{2}e^{(z^{2})})}{(9y^{2})}\\&f_{y}=-\frac{(4x^{2}e^{(z^{2})})}{(9y^{3})}\\&f_{yx}=-\frac{(8xe^{(z^{2})})}{(9y^{3})}\end{align*}\)

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

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot 3^{(3z)}tan(y^{3})e^{(x^{2})})}{5}+ 4x^{4}ycos(z)\\&f_{x}=16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5}\\&f_{xy}=16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}\end{align*}\)

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

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

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

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3y^{2}sin(z^{3})sin(x) - 2\cdot 3^{(y^{2})}sin(x)sin(z)\\&f_{y}=6ysin(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)sin(x)sin(z)\\&f_{yz}=18yz^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)cos(z)sin(x)\\&f_{yzy}=18z^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x)\end{align*}\)

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 partial derivative of the function with respect to x:

\(\displaystyle f_x=2xz^3y+\cos(z)y^2+yz\)

The derivative was found using the following rules:

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

Next, we find the partial derivative of the above function with respect to y:

\(\displaystyle f_{xy}=2xz^3+2y\cos(z)+z\)

We used rules stated above.

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

\(\displaystyle f_{xyz}=6xz^2-2y\sin(z)+1\)

We used all of the rules used previously along with

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)

\(\displaystyle \begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(e^{(4y)}e^{(x)}cos(z^{2} - 3z))}{2})\widehat{i}+(\frac{(4sin(y^{2}))}{z})\widehat{j}+(5e^{(x^{2})}e^{(z)}sin(y))\widehat{k}\\&\text{at the point }(0.98,1.11,2.39)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(\frac{(e^{(4y)}e^{(x)}cos(z^{2} - 3z))}{2})+(\frac{(8ycos(y^{2}))}{z})+(5e^{(x^{2})}e^{(z)}sin(y))\\&divF(0.98,1.11,2.39)=(12.72)+(1.23)+(127.7)=141.66\end{align*}\)

\(\displaystyle \begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(5e^{(y^{2})}e^{(4z)})}{x})\widehat{i}+(\frac{(3ln(x^{2})sin(z^{2}))}{y^{3}})\widehat{j}+(2x^{3}ze^{(y^{4})})\widehat{k}\\&\text{at the point }(1.57,0.68,1.71)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&divF(x,y,z)=(-\frac{(5e^{(y^{2})}e^{(4z)})}{x^{2}})+(-\frac{(9ln(x^{2})sin(z^{2}))}{y^{4}})+(2x^{3}e^{(y^{4})})\\&divF(1.57,0.68,1.71)=(-3009.97)+(-8.19)+(9.58)=-3008.58\end{align*}\)