how to differentiate f(x) = ln ( 1 + √ x / 1 - √ x )

Question

anonymous · Answer

In general, 
$$ \frac{d}{dx} \ln( f(x) ) = \frac{1}{f(x)}\cdot f'(x)$$

anonymous · Answer

how do i solve it? i cant seem to get the correct answer.

anonymous · Answer

Hint: You will have to use the chain rule by finding the derivative of the argument and multiplying it by the derivative that results without the chain rule. Are you aware of how to use the Chain Rule? If not, then I'll provide a solution.

anonymous · Answer

im aware of it. but how do i calculate it?

anonymous · Answer

The derivative of the argument, you mean?

anonymous · Answer

how do i apply chain rule in this equation

anonymous · Answer

Rewrite f(x):
$\Large f(x)=ln\frac{1+\sqrt x}{1-\sqrt x}=ln(1+\sqrt x)-ln(1-\sqrt x) $
now just take the derivatives of each ln separately....

anonymous · Answer

Would you mind showing us your work? What have you found for the derivative of the argument of the natural log?

anonymous · Answer

dy/dx = [ (1/1+√ x) (+ 1/2√x) ] - [ (1/1-√x)  - ( 1/2√ x) ]