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

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Answer 1 · 2016-06-19T03:11:09

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

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

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

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

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

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

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

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