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.

First, we must find the partial derivative of the function with respect to x:

\(\displaystyle f_x=3x^2+y^2\sec^2(xy)\)

Then, we find the partial derivative of this function with respect to x:

\(\displaystyle f_{xx}=6x+2y^3\sec(xy)\tan(xy)\)

To find \(\displaystyle f_{xxz}\) of the function, you take three consecutive partial derivatives:

\(\displaystyle \frac{\partial }{\partial x}(x^2z^2+x^3\cos(z))=2xz^2+3x^2\cos(z)\)

\(\displaystyle \frac{\partial }{\partial x}(2xz^2+3x^2\cos(z))=2z^2+6x\cos(z)\)

\(\displaystyle \frac{\partial }{\partial z}(2z^2+6x\cos(z))=4z-6x\sin(z)\)

To find \(\displaystyle f_{xxz}\) of the function, you must take three consecutive partial derivatives:

\(\displaystyle \frac{\partial }{\partial x}(5x^2z^5+ze^{5x})=10xz^5+5ze^{5x}\)

\(\displaystyle \frac{\partial }{\partial x}(10xz^5+5ze^{5x})=10z^5+25ze^{5x}\)

\(\displaystyle \frac{\partial }{\partial z}(10z^5+25ze^{5x})=50z^4+25e^{5x}\)

To find \(\displaystyle \frac{\partial F}{\partial y}\) we take the derivative of the function \(\displaystyle F(x,y)\) with respect to \(\displaystyle y\) and treat the other variables as constants. 

As such,

\(\displaystyle \frac{\partial }{\partial y}[F(x,y)]=\frac{\partial }{\partial y}[e^x+e^y+cos\,x+sin\,y]\)

yields

\(\displaystyle \frac{\partial F}{\partial y}=e^y+cos\,y\)

To find \(\displaystyle \frac{\partial F}{\partial x}\) we take the derivative of the function \(\displaystyle F(x,y)\) with respect to \(\displaystyle x\) and treat the other variables as constants. 

As such,

\(\displaystyle \frac{\partial }{\partial x}[F(x,y)]=\frac{\partial }{\partial x}[tan\,x-cot\,y+y^2]\)

yields

\(\displaystyle \frac{\partial F}{\partial x}=sec^2\,x\)

To find \(\displaystyle \frac{\partial F}{\partial x}\) we take the derivative of the function \(\displaystyle F(x,y)\) with respect to \(\displaystyle x\) and treat the other variables as constants. 

As such,

\(\displaystyle \frac{\partial }{\partial x}[F(x,y)]=\frac{\partial }{\partial x}[sec\,x+csc\,y]\)

yields

\(\displaystyle \frac{\partial F}{\partial x}=sec(x)tan(x)\)

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.

First, we must find the partial derivative of the function with respect to y:

\(\displaystyle f_y=4y^3z^3+xz^2\)

Next, we find the partial derivative of this function with respect to z:

\(\displaystyle f_{yz}=12y^3z^2+2xz\)

Now, we find the partial derivative of this function with respect to y:

\(\displaystyle f_{yzy}=36y^2z^2\)

Finally, we take the partial derivative of this function with respect to z:

\(\displaystyle f_{yzyz}=72y^2z\)

To find \(\displaystyle f_{xy}\) of the function, you must take two consecutive partial derivatives:

\(\displaystyle \frac{\partial }{\partial x}(x^3\sin(y)+xyz)=3x^2\sin(y)+yz\)

\(\displaystyle \frac{\partial }{\partial y}(3x^2\sin(y)+yz)=3x^2\cos(y)+z\)

To find \(\displaystyle f_{xy}\) of the function, you must take two consecutive partial derivatives:

\(\displaystyle \frac{\partial }{\partial x}(e^x+x\cos(y))=e^x+\cos(y)\)

\(\displaystyle \frac{\partial }{\partial y}(e^x+\cos(y))=-\sin(y)\)

When calculating the partial derivative with respect to the variable \(\displaystyle x\) of a function \(\displaystyle f\) of more than one variable, apply the standard rules for differentiating a function \(\displaystyle f(x)\) of a single variable, and treat the other variables as constants. In this case, we have:

\(\displaystyle f(x,y)=x^2y^2+\frac{x}{x+y}\)

We are being asked to differentiate \(\displaystyle f(x,y)\) with respect to \(\displaystyle x\), so we treat the variable \(\displaystyle y\) as a constant and apply the sum and quotient rules of differentiation to find the partial derivative \(\displaystyle \frac{\partial f}{\partial x}\), as shown:

\(\displaystyle \frac{\partial f}{\partial x}=2x(y^2)+\frac{(1)(x+y)-(1)(x)}{(x+y)^2}\)

\(\displaystyle =2xy^2+\frac{x+y-x}{(x+y)^2}\)

\(\displaystyle =2xy^2+\frac{y}{(x+y)^2}\)

Since \(\displaystyle y\) is treated as a constant, we can factor out \(\displaystyle y\) from the sum of these two terms and simplify the expression:

\(\displaystyle \frac{\partial f}{\partial x}=y(2xy+\frac{1}{(x+y)^2})\)