# Partial Differentiation

• August 4th 2009, 12:26 PM
rologolfer
Partial Differentiation

For an arbitrary function f = f(u, v) show that

1/x(df/dx) + 1/y(df/dy) = 4(df/du)

where u = x^2 + y^2, v = x^2 − y^2,

x not = 0 and y not = 0
• August 4th 2009, 12:37 PM
arbolis
Quote:

Originally Posted by rologolfer

For an arbitrary function f = f(u, v) show that

1/x(df/dx) + 1/y(df/dy) = 4(df/du)

where u = x^2 + y^2, v = x^2 − y^2,

x not = 0 and y not = 0

By the chain rule, $\frac{\partial f}{\partial x}= \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x}+\frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x}$.
Does this help?
• August 4th 2009, 01:00 PM
rologolfer
Thanks a lot, thats great!