Question

Find the first differential of `y= Ln [( x^(1/2)) / ( 1+x^(1/2))]` ?

Answer 1

`dy/dx = 1/[2(x+x^(3/2))]`

Explanation:

I have asumed that the question as written has omitted brackets around the `(1/2)` exponents so that:
`y= Ln [( x^(1/2)) / ( 1+x^(1/2))]`

If this is the case then:

`dy/dx = (1+x^(1/2))/(x^(1/2)` * `d/dx (( x^(1/2)) / ( 1+x^(1/2)))` Standard differential and chain rule

`dy/dx = (1+x^(1/2))/(x^(1/2)` * `[(1+x^(1/2)) * 1/2x^(-1/2) - x^(1/2) * 1/2x^(-1/2)] / (1+x^(1/2))^2` Quotient rule

`= (1+x^(1/2))/(x^(1/2)` * `1/2x^(-1/2) [(1+x^(1/2)-x^(1/2))]/ (1+x^(1/2))^2`

`= (1+x^(1/2))/(x^(1/2)` * `[(1+cancel(x^(1/2))-cancel(x^(1/2)))] / [2x^(1/2)(1+x^(1/2))^2]`

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

`= cancel((1+x^(1/2)))/ [2x (1+x^(1/2))^cancel(2])`

`= 1/[2(x+x^(3/2))]`