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

Explanation

Let `deltax` be a small increment in the value of x.
`:'u` is a function of x, there should be a corresponding increment `deltau` in the value of u. Also, y is a function of u.
`:.` There should be a corresponding increment `deltay` in the value of y.
Consider `(deltay)/(deltax)=(deltay)/(deltau).(deltau)/(deltax)`
Taking `underset(deltaxto0)lim` on both sides, we get
`underset(deltaxto0)lim(deltay)/(deltax)=underset(deltaxto0)lim(deltay)/(deltau).underset(deltaxto0)lim(deltau)/(deltax)`
As `deltaxto0,impliesdeltauto0`.
`:.underset(deltaxto0)lim(deltay)/(deltax)=underset(deltauto0)lim(deltay)/(deltau).underset(deltaxto0)lim(deltau)/(deltax)" "......(i)`
But `underset(deltauto0)lim(deltay)/(deltau)-(deltay)/(deltau)` exists and is finite.
Also `underset(deltaxto0)lim(deltay)/(deltax)=(du)/(dx)` exists and is finite.
`:'` Limits on R.H.S. of (i) exists and are finite, limits on L.H.S. should also exist an be finite.
`:.underset(deltaxto0)lim(deltay)/(deltax).(du)/(dx)`
This is called composite rule or chain rule.