# Thread: help with total differentiation

1. ## 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

2. 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 $w(z)=f(z)\,,\; with\; z=ax+by$, so applying the chain rule:

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

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

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

Tonio