Question

If y = f(u) is a differentiable function of u and u = g(x) is a differentiate function of x such that the composite function y = f[g(x)] is a differentiable function of x then prove that

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

Hence find `dy/dx` if y = log(x² + 5)

Answer 1

Given that y = f(u) and u = g(x).

We assume the function u to be non-constant.

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

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

As δx, δu, δy are small increments in x, u and y respectively such that δx ≠ 0, δu ≠ 0, δy ≠ 0, we have

`(δy)/(δx) = (δy)/(δu) xx (δu)/(δx)`

Taking the limit as δ `->` 0 on both sides,

`lim_(δ -> 0) ((δy)/(δx)) = lim_(δx -> 0) ((δy)/(δu)) xx ((δu)/(δx))`

As δx `->` 0, we get δu `->` 0, since u is a continuous function of x.

∴ `lim_(δx -> 0) ((δy)/(δx)) = lim_(δu -> 0) ((δy)/(δu)) xx lim_(δx -> 0) ((δu)/(δx))`   ...(i)

We have: since y is a differentiable function of u and u is a differentiable function of x

`lim_(δu -> 0) ((δy)/(δu)) = dy/(du)` and `lim_(δx -> 0) ((δu)/(δx)) = (du)/dx`  ...(ii)

From (i) and (ii):

`lim_(δx -> 0) ((δy)/(δx)) = dy/(du) xx (du)/dx`   ...(iii)

The R.H.S. of (iii) exists and is finite, so the L.H.S. must also exist and be finite.

∴ `lim_(δx -> 0) ((δy)/(δx)) = dy/dx`

∴ From (iii): `dy/dx = dy/(du)  xx (du)/dx`

Now, y = log(x² + 5)

Let u = x² + 5,

∴ y = log u

∴ `(du)/dx` = 2x

and `dy/(du) = 1/u`

∴ `dy/dx = dy/(du) xx (du)/dx`

= `1/u(2x)`

= `(2x)/(x^2 + 5)`