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 ux = 2y/z, uy = 2x/z, and uz = -2xy/z2.
Let's solve this step by step.
u = 2xy/z
We have to find the values of ux, uy, uz. For this, we have to do the partial differentiation with respect to x, y, and z respectively.
Let's find ux first, here y and z are constants.
ux = du/dx = 2y/z
Let's find uy now, here x and z are constants.
uy = du/dy = 2x/z
Let's find uz first, here x and y are constants.
uz = du/dz = -2xy/z2
Hence, the first partial derivatives of the function u = 2xy/z are ux = 2y/z, uy = 2x/z, and uz = -2xy/z2.