Question

how do I find the second derivative of ln(x/x-1)?

Answer 1

use chain rule

Answer 2

\[\frac{d}{dx}(\ln(\frac{x}{x-1}))= (\frac{x}{x-1})' \cdot \frac{1}{\frac{x}{x-1}}\]

Answer 3

i got that the first derivative was 1/x - 1/x-1

Answer 4

or I have an easier way

Answer 5

oh and you used the easier way

Answer 6

ln(x/(x-1))=ln(x)-ln(x-1)
derivative is 1/x - 1/(x-1)
gj

Answer 7

the second derivative is the derivative of (1/x) - (1/(x-1))

Answer 8

\[f(x)=\ln(\frac{x}{x-1}) =\ln(x)-\ln(x-1) \\ f'(x)=\frac{1}{x}-\frac{1}{x-1}=x^{-1}-(x-1)^{-1} \\ f''(x)=?\]

Answer 9

you could have used quotient rule

Answer 10

or you could just use chain rule

Answer 11

oh that makes more since! thank you

Answer 12

np