Chain rule in partial derivative

[bigwuying]

There is a theorem in partial derivative
If x= x(t) , y= y(t), z= z(t) are differentiable at t_{0}, and if w= f(x,y,z) is differentiable at the point (x,y,z)=(x(t),y(t),z(t)),then w=f(x(t),y(t),z(t)) is differentiable at t and
\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} + \frac{\partial w}{\partial z} \frac{dz}{dt}
where the ordinary derivatives are evaluated at t and the partial derivatives are evaluated at (x,y,z)

My question are
1. What is the difference between dx and\partial x?
2. i don't know why i am not allowed to eliminate the dx and \partial x,also the same for y and z to get 3\frac{\partial w}{dt}

 

[HallsofIvy]

The problem appears to be that you do not understand the derivative itself. Neither "\partial x" nor "dx" have any meaning alone. And, when using the "ordinary" chain rule- (df/dx)(dx/dy)= df/dy- you are NOT "canceling" the "dx"s.

 

[bigwuying]

Yes, you are right. i only know how to use chain rule, but i don't know what it is actually.
When i start to learn Partial Derivative, i find it difficult to understand why dw/dt appear 3 term, and why do dx and dx not be eliminate.

To understand this problem ,i have read my text and start confusing

The definition of partial d of z with repect to x at x_{0},y_{0} is
f_{x}(x_{0},y_{0})=lim_{x->x0}\frac{f(x,y0)-f(x0-y0)}{x-x0}= \frac {\partialf}{\partialx}

Consider a part of the chain rule, let say : dw/dt= \frac{\partial w}{\partial x} \frac{dx}{dt}

it is equivalent to \left[lim_{x->x0}\frac{f(x,y0,z0)-f(x0,y0,z0)}{x-x0}\right] \left[lim_{t->t0}\frac{x(t)-x(t0)}{t-t0}\right]

lim_{x->x0} x-x_{0} isn't the same as lim_{t->t_{0}} x(t)-x(t_{0}) ??
So i get confusing why i can't eliminate them. or i cannot treat them as a fraction??