Question

lf y = f(u) is a differentiable function of u and u = g(x) is a differentiable 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 `d/dx[log(x^5 + 4)]`.

Answer 1

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

We assume that u is not a constant function.

Let δu and δy be the increments in u and y respectively, corresponding to the increment δx in x.

Now, y is a differentiable function of u and u is a differentiable function of x.

∴ `dy/(du) = lim_(δu -> 0) (δy)/(δu)` and `(du)/(dx) = lim_(δx -> 0) (δu)/(δx)`     ...(1)

Also, `lim_(δx -> 0) δu = lim_(δu -> 0) ((δu)/(δx).δx)`

= `(lim_(δx -> 0) (δu)/(δx))(lim_(δx -> 0) δx)`

= `(du)/dx xx 0`

= 0

This means that as `δx → 0, δu → 0`   ...(2)

Now, `(δy)/(δx) = (δy)/(δu) xx (δu)/(δx)`   ...[∵ δu ≠ 0] 

Taking limits as `δx → 0`, we get

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

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

= `lim_(δu -> 0) (δy)/(δu) xx lim_(δx -> 0)(δu)/(δx)`  ...[By (2)]

Now, both the limits on RHS exist  ...[By (1)]

∴ `lim_(δx -> 0) (δy)/(δx)` exists and is equal to `dy/dx`  

∴ y is differentiable function of x and

`dy/dx = dy/(du) xx (du)/dx`.  ...[By (1)]

To find `bb(d/dx[log(x^5 + 4)])`:

Let y = log(x⁵ + 4)

Put u = x⁵ + 4

Then y = log u

∴ `dy/(du) = d/(du)(log u)`

= `1/u`

= `1/(x^5 + 4)`

and `(du)/dx = d/dx(x^5 + 4)`

= 5x⁴ + 0

= 5x⁴

∴ `dy/dx = dy/(du) xx (du)/dx`

= `1/(x^5 + 4) xx 5x^4`

= `(5x^4)/(x^5 + 4)`