Question

Find ∂f∂x, ∂f∂y, and ∂f∂z.f(x, y, z) = 6xy + 6xz − 6yz

Answer 1

The given function is

$f(x,y,z)=6xy+6xz-6yz$

To find df/dx we will equate the term that without x by 0

$\begin{gathered} \frac{df}{dx}=6y+6z-0 \\ \frac{df}{dx}=6y+6z \\ \frac{df}{dx}=6(y+z) \end{gathered}$

To find df/dy we will equate the term that has not y by 0

$\begin{gathered} \frac{df}{dy}=6x+0-6z \\ \frac{df}{dy}=6x-6z \\ \frac{df}{dy}=6(x-z) \end{gathered}$

To find df/dz we will consider the term that has no z as a constant, then its differentiation is 0

$\begin{gathered} \frac{df}{dz}=0+6x-6y \\ \frac{df}{dz}=6x-6y \\ \frac{df}{dz}=6(x-y) \end{gathered}$

We need to find them at (0, -1, 1)

That means x = 0, y = -1, z = 1

$\begin{gathered} \frac{df}{dx}(0,-1,1)=6(-1+1) \\ \frac{df}{dx}(0,-1,1)=6(0) \\ \frac{df}{dx}(0,-1,1)=0 \end{gathered}$$\begin{gathered} \frac{df}{dy}(0,-1,1)=6(0-1) \\ \frac{df}{dy}(0,-1,1)=6(-1) \\ \frac{df}{dy}(0,-1,1)=-6 \end{gathered}$$\begin{gathered} \frac{df}{dz}(0,-1,1)=6(0--1) \\ \frac{df}{dz}(0,-1,1)=6(0+1) \\ \frac{df}{dz}(0,-1,1)=6 \end{gathered}$