Question

If y = f(u) is a differentiable function of u and u = g(x) is 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(1/sqrtsinx)`.

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_(δx -> 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)`

= `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: `d/dx(1/sqrtsinx)`

Let y = `1/sqrtsinx`

Put u = sin x. Then y = `1/sqrtu`

∴ `(dy)/(du) = d/(du)(1/sqrtu)`

= `d/(du) (u^(-1/2))`

= `-1/2 u^(-3/2)`

= `-1/2 (sin x)^(-3/2)`

and `(du)/(dx) = d/(dx) (sin x) = cos x`

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

= `-1/2 (sin x)^(-3/2) · cos x`