Question

What is the derivative of `sqrt(x - 1)/sqrtx`?

Answer 1

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

Explanation:

To find the derivative of `sqrt(x-1)/sqrt(x)` we can try the following
`sqrt(x-1)/sqrt(x) = sqrt((x-1)/x)`
on simplifying further

`=sqrt(1-1/x)`

Let us use chain rule.

Let `y=sqrt(u)` and `u=1-1/x`

By chain rule

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

`y=sqrt(u)`
`dy/(du) = 1/(2sqrt(u))`
Substituting back for `u` we get
`dy/(du) = 1/(2sqrt(1-1/x))`

Now we shall find our `(du)/dx`

`u=1-1/x`
`(du)/dx = 0- (-1/x^2)`
`(du)/dx = 1/x^2`

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

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