Can you find the derivative of ln

f(x)= ln ((x-1)/(x+1)

also did i do this one correctly?
f(x)= e^(x^2 -x -6)
dy/dx= (2x-1)e^(x^2 -x -6)

f(x) = ln[(x - 1)/(x + 1)] = ln(x - 1) - ln(x + 1)

f '(x) = 1/(x - 1) - 1/(x + 1) = 2/(x² - 1)

You did differentiate the other function correctly.

yuup

d/dx (ln (x)) = 1/x

so when ever you take the derivative of a ln just remember to take its "x" and flip it :) [when its x by itself]
but if you have more than just x (like x^2 or 2x-1) you're going to use the chain rule

for the first one, expand the ((x-1)/(x+1)) using log rules

f(x)= ln (x-1) - ln (x+1)

using the chain rule
take the derivative of the outside and multiplied by the derivative of the inside

so the derivative of the outside is the ln, which we know will take shape as the reciprocal of the "x" portion

so we have

1/(x-1) - 1/(x+1)

then the inside would be "x" part, in this case
x-1 and x+1

(x-1)' = 1

(x+1)'= 1

multiplying each of the ln by its inner part

1*(1/x-1) - 1*(1/x+1)

and combine them giving the fraction a common denominator :)

it's a little tedious going through the trouble of taking the derivative of the inner just find that it's just 1 but it doesn't always workout that way, so i emphasized that point a bit.

and your second prob is right :)

hope this helped