Question

Let f(x) = ln(1 + (1/x)). Show that f'(x) = (x^2 - 1)/ (x^3 +x).

Answer 1

for starters
\[1+\frac{1}{x}=\frac{x+1}{x}\] and
\[\ln\left(\frac{x+1}{x}\right)=\ln(x+1)-\ln(x)\] therefore taking the derivative should be relatively painless

Answer 2

I get the first part, but how does the ln( (x+1)/x) = ln(x+1)-ln(x) make sense?

Answer 3

It is one of the logarithmic identities:

log(A/B) = log(A) - log(B)
log(A * B) = log(A) + log(B)
etc...

(same if you replace log with ln)

Answer 4

yes BUT that doesn't help me find the derivative at all. the derivative is (x^2 -1)/(x^3 -1). Using the identity leads me to a dead end.

Answer 5

\[f(x)=\ln\left(\frac{x+1}{x}\right)=\ln(x+1)-\ln(x)\]
\[f'(x)=\frac{1}{x+1}-\frac{1}{x}\] is a start

Answer 6

ooh i think you wrote something different above

Answer 7

\[\frac{x^2-1}{x^3-1}=\frac{(x+1)(x-1)}{(x-1)(x^2+x+1)}=\frac{x+1}{x^2+x+1}\]

Answer 8

Actually I figured it out! Thanks for your help!