Question

If y = f(u) is a 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) * (du)/(dx)`.

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

Answer 1

Let δx be a small increment in x.

Let δy and δu be the corresponding increments in y and u, respectively.

As δx → 0, δy → 0, δu → 0.

As u is differentiable function, it is continuous.

Consider the incrementary ratio `(deltay)/(deltax)`

We have, `(deltay)/(deltax) = (deltay)/(deltau) xx (deltau)/(deltax)`

Taking limit as δx → 0, on both sides,

`lim_(deltax -> 0)(deltay)/(deltax) = lim_(deltax -> 0)((deltay)/(deltau) xx (deltau)/(deltax))`

`lim_(deltax -> 0)(deltay)/(deltax) = lim_(deltau -> 0)(deltay)/(deltau) xx lim_(deltax -> 0)(deltau)/(deltax)`   ...(1)

Since y is a differentiable function of u, `lim_(deltau -> 0)(deltay)/(deltau)` exists and  `lim_(deltau -> 0)(deltay)/(deltax)` exists as u is a differentiable function of x.

Hence, R.H.S. of (1) exists

Now `lim_(deltau -> 0)(deltay)/(deltau) = (dy)/(du) and lim_(deltau -> 0)(deltau)/(deltax) = (du)/(dx)`

`lim_(deltax -> 0)(deltay)/(deltax) = (dy)/(du) xx (du)/(dx)`

Since R.H.S. exists, L.H.S. of (1) also exists and 

`lim_(deltax -> 0)(deltay)/(deltax) = (dy)/(dx)`

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