Question

Find `dy/dx` if, y = `sqrt(x + 1/x)`

Answer 1

y = `sqrt(x + 1/x)`

Let u = `x + 1/x`

∴ y = `sqrt u`

Differentiating both sides w.r.t. u, we get

`dy/(du) = d/(du) (sqrt u) = 1/(2sqrt u)`

u = `x + 1/x`

Differentiating both sides w.r.t. x, we get

`(du)/dx = d/dx (x + 1/x) = 1 - 1/x^2`

By chain rule, we get

`dy/dx = dy/(du) xx (du)/dx = 1/(2sqrt u) xx (1 - 1/x^2)`

= `1/(2sqrt(x + 1/x)) (1 - 1/x^2)`

∴ `dy/dx = 1/2 (x + 1/x)^(-1/2)(1 - 1/x^2)`

Alternate Method:

y = `sqrt(x + 1/x)`

∴ y = `(x + 1/x)^(1/2)`

Differentiating both sides w.r.t.x, we get

`dy/dx = d/dx [(x + 1/x)^(1/2)]`

= `1/2 (x + 1/x)^(-1/2) * d/dx (1 + 1/x)`

`dy/dx = 1/2 (x + 1/x)^(-1/2) (1 - 1/x^2)`