Question

find ( f_x(x,y,z), f_y(x,y,z), f_z(x,y,z) ), and ( f_{yx}(x,y,z) ) for the following.
( f(x,y,z)=3x^{3}+2xy - 4z^{4} )
( f_x(x,y,z)= )

Explanation:

Step1: Find \(f_x(x,y,z)\)

Differentiate \(f(x,y,z) = 3x^{3}+2xy - 4z^{4}\) with respect to \(x\) (treating \(y\) and \(z\) as constants).
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\) and \(\frac{d}{dx}(ax)=a\) (where \(a\) is a constant), we have:
\(f_x(x,y,z)=\frac{\partial}{\partial x}(3x^{3}+2xy - 4z^{4})=3\times3x^{2}+2y-0 = 9x^{2}+2y\)

Step2: Find \(f_y(x,y,z)\)

Differentiate \(f(x,y,z) = 3x^{3}+2xy - 4z^{4}\) with respect to \(y\) (treating \(x\) and \(z\) as constants).
\(f_y(x,y,z)=\frac{\partial}{\partial y}(3x^{3}+2xy - 4z^{4})=0 + 2x-0=2x\)

Step3: Find \(f_z(x,y,z)\)

Differentiate \(f(x,y,z) = 3x^{3}+2xy - 4z^{4}\) with respect to \(z\) (treating \(x\) and \(y\) as constants).
Using the power rule \(\frac{d}{dz}(z^{n})=nz^{n - 1}\), we get \(f_z(x,y,z)=\frac{\partial}{\partial z}(3x^{3}+2xy - 4z^{4})=0+0-4\times4z^{3}=-16z^{3}\)

Step4: Find \(f_{yx}(x,y,z)\)

First, we know \(f_y(x,y,z) = 2x\). Then differentiate \(f_y(x,y,z)\) with respect to \(x\) (treating \(y\) and \(z\) as constants).
\(f_{yx}(x,y,z)=\frac{\partial}{\partial x}(2x)=2\)

Answer:

\(f_x(x,y,z)=9x^{2}+2y\), \(f_y(x,y,z)=2x\), \(f_z(x,y,z)=-16z^{3}\), \(f_{yx}(x,y,z) = 2\)