\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2\cdot4^{(3y)}x^2cos{(z)})}}{7}\\&f_{x}=\frac{{(4\cdot4^{(3y)}xcos{(z)})}}{7}\\&f_{xx}=\frac{{(4\cdot4^{(3y)}cos{(z)})}}{7}\\&f_{xxz}=\frac{-{(4\cdot4^{(3y)}sin{(z)})}}{7}\\&f_{xxzy}=\frac{-{(12\cdot4^{(3y)}ln{(4)}sin{(z)})}}{7}\end{align*}\)

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=2\cdot3^{(3x)}\cdot3^ysin{(z^2)}\\&f_{x}=6\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}\\&f_{xx}=18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2\end{align*}\)

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=2\cdot3^{(3x)}\cdot3^ysin{(z^2)}\\&f_{x}=6\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}\\&f_{xx}=18\cdot3^{(3x)}\cdot3^ysin{(z^2)}ln{(3)}^2\end{align*}\)

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(3^{(y^2)}sin{(4z)})}}{{(4x^2)}}\\&f_{x}=\frac{-{(3^{(y^2)}sin{(4z)})}}{{(2x^3)}}\\&f_{xx}=\frac{{(3\cdot3^{(y^2)}sin{(4z)})}}{{(2x^4)}}\end{align*}\)

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

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3z^3sin{(y^3)}ln{(x)} -\frac{ {(2\cdot2^xy^2cos{(4z)})}}{9}\\&f_{y}=9y^2z^3cos{(y^3)}ln{(x)} -\frac{ {(4\cdot2^xycos{(4z)})}}{9}\\&f_{yy}=18yz^3cos{(y^3)}ln{(x)} - 27y^4z^3sin{(y^3)}ln{(x)} -\frac{ {(4\cdot2^xcos{(4z)})}}{9}\end{align*}\)

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=5y^3e^{(z^2)} -\frac{ {(sin{(3z)}e^{(3x)}e^{(y^2)})}}{5}\\&f_{y}=15y^2e^{(z^2)} -\frac{ {(2ysin{(3z)}e^{(3x)}e^{(y^2)})}}{5}\\&f_{yy}=30ye^{(z^2)} -\frac{ {(2sin{(3z)}e^{(3x)}e^{(y^2)})}}{5} -\frac{ {(4y^2sin{(3z)}e^{(3x)}e^{(y^2)})}}{5}\end{align*}\)

This derivative is most easily done using logarithmic differentiation.  Consider the following

\(\displaystyle \begin{align*} f(x,y)&=x^{y+x} \\ \ln f&= \ln x \cdot (y+x) \\ \ln f&= \ln x \cdot y + \ln x \cdot x \\ \frac{f_y}{f}&= \ln x\\ f_y &= f \cdot \ln x = \ln x \cdot x^{y+x} \end{align*}\)

While taking a derivative with respect to only one variable, the other variables are treated as constants.  Consider the following

\(\displaystyle \begin{align*} \frac{\partial f}{\partial z}&= \frac{\partial}{\partial z} \left( -\frac{x^2+y^2}{\cos z}\right )\\ &= -\left(x^2+y^2 \right )\frac{\partial }{\partial z} \left(\cos z \right )^{-1}\\ &=- \left(x^2+y^2 \right )\frac{-\sin z}{\cos ^2 z} \\ \frac{\partial f}{\partial z}&= \frac{\left(x^2+y^2 \right ) \tan z}{\cos z} \end{align*}\)

\(\displaystyle \begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{{(2^{{({4x)}}}z^{2}e^{{({4y)}}})}}{5 }+\frac{ {(3e^{{({x^{2})}}}e^{{({y^{2})}}})}}{z}\\&f_{z}=\frac{{(2\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }-\frac{ {(3e^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2}}\\&f_{zy}=\frac{{(8\cdot2^{{({4x)}}}ze^{{({4y)}}})}}{5 }-\frac{ {(6ye^{{({x^{2})}}}e^{{({y^{2})}}})}}{z^{2}}\end{align*}\)