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

Answer 1

#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))]#

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Answer 2

Christopher Allers

To find the first derivative of ( y = \ln\left(\frac{x^\frac{1}{2}}{1+x^\frac{1}{2}}\right) ), use the chain rule and the quotient rule. The first derivative is:

[ \frac{dy}{dx} = \frac{1}{2}\left(\frac{1}{x^\frac{1}{2}} - \frac{1}{1+x^\frac{1}{2}}\right) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.