How to Find the Derivative of f(x)=ln[x/(x-1)]?

lreichardt
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How do I take the derivative of

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

Thanks!
 
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Welcome to PF!

Hi lreichardt! Welcome to PF! :smile:

Just use the chain rule: derivative of ln, times the derivative of x/(x-1). :smile:
 
so, (x/1)(quotient rule of x/(x-1)? =

(x/1)(-x/x-1)=-x^2/x-1
 
ooh dear …

better go the easy way … simplify it first …

ln(x/(x-1)) = ln(x) - ln(x-1) … now differentiate! :smile:
 
Chain rule = The difference of the derivative of the "Outer" and derivative of the "Inner", put simply.
 
Thanks all, figured it out.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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