# Manipulating differential equation

#### Meta

v(x)p(x) = d/dx (D(x)p(x)) can be rearranged to 1/(D(x)p d/dx(D(x)p(x)) = d/dx ln(D(x)p(x)) where d/dx are partial differentials. Can anyone explain how this works to me? Thanks

#### topsquark

Math Team
v(x)p(x) = d/dx (D(x)p(x)) can be rearranged to 1/(D(x)p d/dx(D(x)p(x)) = d/dx ln(D(x)p(x)) where d/dx are partial differentials. Can anyone explain how this works to me? Thanks
I haven't played with it yet, but where did the v(x) go?

-Dan

#### Meta

Apologies, this was just the manipulation of the RHS. I see what they've done now - multiplied through by 1/(D(x)p(x)) to give RHS of 1/(D(x)p(x)) d/dx (D(x)p(x)) then applied the chain rule to give RHS = ln d/dx (D(x)p(x)). Then this equals v(x)p(x)/D(x)p(x) = v(x)/D(x). Thanks