# help with total differentiation

• Oct 29th 2009, 02:50 AM
pepsi
help with total differentiation
Hello,

If w=f(ax+by) I am asked to show -please read w[x] as partial derivative fo w w.r.t.x- that: bw[x]-aw[y]=0 this tells me that the difference of a ccahnge in the coordinates of a points on on a curve in different directiosn will be zero (though pls dont be too harsh if I have - which i probably have - got the whole geometry thing wrong

i am given as a hint to let ax +by = z

i can totally differentiate this : dz = adx + bdy

and totally differentiated w: dw = w[x]dx + w[y]dy but don't know hyow to come up with the desried result.

pls help
• Oct 29th 2009, 03:41 AM
tonio
Quote:

Originally Posted by pepsi
Hello,

If w=f(ax+by) I am asked to show -please read w[x] as partial derivative fo w w.r.t.x- that: bw[x]-aw[y]=0 this tells me that the difference of a ccahnge in the coordinates of a points on on a curve in different directiosn will be zero (though pls dont be too harsh if I have - which i probably have - got the whole geometry thing wrong

i am given as a hint to let ax +by = z

i can totally differentiate this : dz = adx + bdy

and totally differentiated w: dw = w[x]dx + w[y]dy but don't know hyow to come up with the desried result.

pls help

You have that $\displaystyle w(z)=f(z)\,,\; with\; z=ax+by$, so applying the chain rule:

$\displaystyle w_x=\frac{\partial f}{\partial z}\,\frac{\partial z}{\partial x}=a\cdot \frac{\partial f}{\partial z}$

$\displaystyle w_y=\frac{\partial f}{\partial z}\,\frac{\partial z}{\partial y}=b\cdot \frac{\partial f}{\partial z}$

Well, now multiply the first one above by b and the second one by a, substract and get zero...

Tonio