Unizor - Partial Derivatives - ∂²/∂x ∂y = ∂²/∂y ∂x

Notes to a video lecture on http://www.unizor.com

Partial Derivatives -
∂²/∂x ∂y = ∂²/∂y ∂x

In the following examples compare ∂²z/∂x ∂y and ∂²z/∂y ∂x.
They should be identical.

Example 1

Let z=√x·y
Then
∂z/∂x = y/(2√x·y)
∂²z/∂y ∂x = 1/(4√x·y)
∂z/∂y = x/(2√x·y)
∂²z/∂x ∂y = 1/(4√x·y)

Example 2

Let z=e^x·y
Then
∂z/∂x = y·e^x·y
∂²z/∂y ∂x = (x·y+1)·e^x·y
∂z/∂y = x·e^x·y
∂²z/∂x ∂y = (x·y+1)·e^x·y

Example 3

Let z=1/(x²+y²)
Then
∂z/∂x = −2x/(x²+y²)²
∂²z/∂y ∂x = 8x·y/(x²+y²)³
∂z/∂y = −2y/(x²+y²)²
∂²z/∂x ∂y = 8x·y/(x²+y²)³

Example 4

Let z=sin(x)/y²
Then
∂z/∂x = cos(x)/y²
∂²z/∂y ∂x = −2cos(x)/y³
∂z/∂y = −2sin(x)/y³
∂²z/∂x ∂y = −2cos(x)/y³

Example 5

Let z=arctan(x√y)
Then
∂z/∂x = √y/(1+x²·y)
∂²z/∂y ∂x = (1−x²·y)/2√y(1+x²·y)²
∂z/∂y = x/[2√y(1+x²·y)]
∂²z/∂x ∂y = (1−x²·y)/2√y(1+x²·y)²

Example 6

Let z=y^x
Then
∂z/∂x = y^x·ln(y)
∂²z/∂y ∂x = y^x−1·(x·ln(y)+1)
∂z/∂y = x·y^x−1
∂²z/∂x ∂y = y^x−1·(x·ln(y)+1)