Q. 36 - Calculus

Question

Find the first-order partial derivatives for the functions in Exercises 27–36.  

$g\left(x,y,z\right)=2xy+2xz+2yz$

Step-by-Step Solution

Answer

The first-order partial derivatives are $\frac{\partial g}{\partial x}=2y+2z, \frac{\partial g}{\partial y}=2x+2z, \frac{\partial g}{\partial z}=2x+2y.$

Step 1: Given

The given function is: $g\left(x,y,z\right)=2xy+2xz+2yz$

Step 2: To find

We have to find the first-order partial derivatives.  

Step 3: Calculation

$$\begin{gathered} g\left(x,y,z\right)=2xy+2xz+2yz \\ \frac{\partial g}{\partial x}=\frac{\partial }{\partial x}\left(2xy+2xz+2yz\right) \\ \frac{\partial g}{\partial x}=2y+2z \\ g\left(x,y,z\right)=2xy+2xz+2yz \\ \frac{\partial g}{\partial y}=\frac{\partial }{\partial y}\left(2xy+2xz+2yz\right) \\ \frac{\partial g}{\partial y}=2x+2z \\ g\left(x,y,z\right)=2xy+2xz+2yz \\ \frac{\partial g}{\partial z}=\frac{\partial }{\partial z}\left(2xy+2xz+2yz\right) \\ \frac{\partial g}{\partial z}=2x+2y \\ \text{Hence}, \frac{\partial g}{\partial x}=2y+2z, \frac{\partial g}{\partial y}=2x+2z, \frac{\partial g}{\partial z}=2x+2y. \end{gathered}$$