To find the partial derivative \(\displaystyle f_x\) of \(\displaystyle f(x,y)=x^2\sin e^y\), we take its derivative with respect to \(\displaystyle x\) while holding \(\displaystyle y\) constant.

So we get

\(\displaystyle \\f_x=\frac{\partial}{\partial x}f(x,y) \\f_x=\frac{\partial}{\partial x}x^2\sin e^y \\f_x=\sin e^y \cdot \frac{\partial}{\partial x}x^2 \\f_x=2x\sin e^y\)

To find the partial derivative \(\displaystyle f_y\) of \(\displaystyle f(x,y)=x^2\sin e^y\), we take its derivative with respect to \(\displaystyle y\) while holding \(\displaystyle x\) constant.

So we get

\(\displaystyle \\f_y=\frac{\partial}{\partial y}f(x,y) \\ \\f_y=\frac{\partial}{\partial y}x^2\sin e^y \\ \\f_y=x^2 \cdot \frac{\partial}{\partial y}\sin e^y \\ \\ f_y=2x\cos e^y\cdot e^y \\ \\f_y=x^2e^y\cos e^y\)

Given the function \(\displaystyle f(x,y,z)=z^3e^{5x+y^4}\), we can find the partial derivative \(\displaystyle f_x\) by taking its derivative with respect to \(\displaystyle x\) while holding \(\displaystyle y,z\) constant.

So we get

\(\displaystyle \\f_x=\frac{\partial}{\partial x}f(x,y,z) \\ \\f_x=\frac{\partial}{\partial x}z^3e^{5x+y^4} \\ \\f_x=z^3\frac{\partial}{\partial x}e^{5x+y^4} \\ \\f_x=z^3e^{5x+y^4}\frac{\partial}{\partial x}(5x+y^4)\)

\(\displaystyle =5z^3e^{5x+y^4}\)

Given the function \(\displaystyle f(x,y,z)=z^3e^{5x+y^4}\), we can find the partial derivative \(\displaystyle f_y\) by taking its derivative with respect to \(\displaystyle y\) while holding \(\displaystyle x,z\) constant.

So we get

\(\displaystyle \\f_y=\frac{\partial}{\partial y}f(x,y,z) \\ \\f_y=\frac{\partial}{\partial y}z^3e^{5x+y^4} \\ \\f_y=z^3\frac{\partial}{\partial y}e^{5x+y^4} \\ \\f_y=z^3e^{5x+y^4}\frac{\partial}{\partial y}(5x+y^4)\)

\(\displaystyle =4y^3z^3e^{5x+y^4}\)

Given the function \(\displaystyle f(x,y,z)=z^3e^{5x+y^4}\), we can find the partial derivative \(\displaystyle f_z\) by taking its derivative with respect to \(\displaystyle z\) while holding \(\displaystyle x,y\) constant.

So we get

\(\displaystyle \\f_z=\frac{\partial}{\partial z}f(x,y,z) \\ \\f_z=\frac{\partial}{\partial z}z^3e^{5x+y^4} \\ \\f_z=e^{5x+y^4}\frac{\partial}{\partial y}z^3 \\ \\ f_z=3z^2e^{5x+y^4}\)

To find the partial derivative \(\displaystyle f_x\) of the function \(\displaystyle f(x,y,z)=y\ln(x^2+z^5)\), we take its derivative with respect to \(\displaystyle x\) while holding \(\displaystyle y,z\) constant.

We use the chain rule to get

\(\displaystyle \\f_x=\frac{\partial}{\partial x}f(x,y,z) \\ \\f_x=\frac{\partial}{\partial x}y\ln(x^2+z^5) \\ \\f_x=y\frac{\partial}{\partial x}\ln(x^2+z^5) \\ \\f_x=y\cdot \frac{\frac{\partial}{\partial x}(x^2+z^5)}{x^2+z^5}\)

\(\displaystyle =\frac{2xy}{x^2+z^5}\)

To find the partial derivative \(\displaystyle f_y\) of the function \(\displaystyle f(x,y,z)=y\ln(x^2+z^5)\), we take its derivative with respect to \(\displaystyle y\) while holding \(\displaystyle x,z\) constant.

We get

\(\displaystyle \\f_y=\frac{\partial}{\partial y}f(x,y,z) \\ \\f_y=\frac{\partial}{\partial y}y\ln(x^2+z^5) \\ \\f_y=\ln(x^2+z^5)\frac{\partial}{\partial y}y \\ \\f_y=\ln(x^2+z^5)\)

To find the partial derivative \(\displaystyle f_z\) of the function \(\displaystyle f(x,y,z)=y\ln(x^2+z^5)\), we take its derivative with respect to \(\displaystyle z\) while holding \(\displaystyle x,y\) constant.

We use the chain rule to get

\(\displaystyle \\f_z=\frac{\partial}{\partial z}f(x,y,z) \\ \\f_z=\frac{\partial}{\partial z}y\ln(x^2+z^5) \\ \\f_z=y\frac{\partial}{\partial z}\ln(x^2+z^5) \\ \\f_z=y\cdot \frac{\frac{\partial}{\partial z}(x^2+z^5)}{x^2+z^5}\)

\(\displaystyle =\frac{5yz^4}{x^2+z^5}\)

Note that for this problem, we're told to take the derivative with respect to one particular variable. This is known as taking a partial derivative; often it is denoted with the Greek character delta, \(\displaystyle \delta\), or by the subscript of the variable being considered such as \(\displaystyle f_x\) or \(\displaystyle f_y\).

For a problem like this, where we presume all variables are independent of each other, we need only consider the variable that we're taking the derivative of the function with respect to; all other variables can be treated as constants.

Taking the partial derivative of \(\displaystyle f(x,y)=x^y\) at \(\displaystyle (2,3)\)

We find:

\(\displaystyle f_x= yx^{y-1}\rightarrow3(2^2)=12\)

Note that for this problem, we're told to take the derivative with respect to one particular variable. This is known as taking a partial derivative; often it is denoted with the Greek character delta, \(\displaystyle \delta\), or by the subscript of the variable being considered such as \(\displaystyle f_x\) or \(\displaystyle f_y\).

For a problem like this, where we presume all variables are independent of each other, we need only consider the variable that we're taking the derivative of the function with respect to; all other variables can be treated as constants.

Taking the partial derivative of \(\displaystyle f(x,y)=x^3(y+5)\) at \(\displaystyle (2,1)\)

We find:

\(\displaystyle f_x= 3x^2(y+5)\rightarrow12(6)=72\)