#### Question

If y = f (u) is a differential function of u and u = g(x) is a differential function of x, then prove that y = f [g(x)] is a differential function of x and `dy/dx=dy/(du) xx (du)/dx`

#### Solution

Let δx be a small increment in the value of x.

Since u is a function of x, there should be a corresponding increment u in the value of u.

Also y is a function of u.

there should be a corresponding increment δy in the value of y.

Consider,

`(deltay)/(deltax)=(deltay)/(deltau)xx(deltau)/(deltax)`

Taking `lim_(deltax->0)` on both sides, we get

`lim_(deltax->0)(deltay)/(deltax)=lim_(deltax->0)(deltay)/(deltau)xxlim_(deltax->0)(deltau)/(deltax) [because deltax->0,deltau->0]`

`lim_(deltax->0)(deltay)/(deltax)=lim_(deltau->0)(deltay)/(deltau)xxlim_(deltax->0)(deltau)/(deltax) ....(i)`

`But lim_(deltax->0)(deltau)/(deltax)=(du)/(dx) "exists and is finite."`

`also lim_(deltau->0)(deltay)/(deltau)=(dy)/(du)`

limits on R.H.S. of (i) exist and are finite.

Hence, limits on L.H.S. should also exist and be finite

`lim_(deltax->0)(deltay)/(deltax)=dy/dx " exisit and finite"`

`(dy)/(dx)=(dy)/(du)xx(du)/(dx)`