Deriving dy/dx of y=logε((1+√x)(/1-√x))

[BonBon101]

Homework Statement

if y=logε((1 + √x)(/1- √x )) , show than

dy/dx = (1/(√x(1-x))

The Attempt at a Solution

y=logε((1 + √x)(/1- √x ))

y=logε(1+√x)-logε(1-√x)

dy/dx = 1-1/(1+√x) + 1/(1-√x)

i don't no what to do after this or if I am even going in the right direction!?

 

[annoymage]

how you differentiate y=log_e(1+$$\sqrt{x}$$) ?

 

[cristo]

why not? for example if y is given as
$$y=ln(x^2)$$
then
$$\frac{dy}{dx}=\frac{1}{x^2}*2*x=\frac{2}{x}$$

anonymage posted a question, as one would do if tutoring a student face to face. He wasn't saying that you can't differentiate a logarithm, but asking the OP if he knew how to differentiate a logarithm. That is the point of the homework forums here: to help guide a student to the solution to a problem.