Find the First Partial Derivatives of the Function. u = 2xy/z

We will be using the concept of differentiation to solve this.

Answer: The First Partial Derivatives of the Function u = 2xy/z are u_x = 2y/z, u_y = 2x/z, and u_z = -2xy/z².

Let's solve this step by step.

Explanation: 

u = 2xy/z

We have to find the values of u_x, u_y, u_z.  For this, we have to do the partial differentiation with respect to x, y, and z respectively.

Let's find u_x first, here y and z are constants.

u_x = du/dx = 2y/z 

Let's find u_y now, here x and z are constants.

u_y = du/dy = 2x/z

Let's find u_z first, here x and y are constants.

u_z = du/dz = -2xy/z²  

Hence, the first partial derivatives of the function u = 2xy/z are u_x = 2y/z, u_y = 2x/z, and u_z = -2xy/z².