How do you differentiate #y= ln (x/(x-1))#?

Question

Luke Alvarez · Accepted Answer

#(-1)/(x(x-1)# Using #color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(lnf(x))=1/(f(x)))color(white)(a/a)|)))# combined with the #color(blue)" chain rule"# #color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx[f(g(x))]=f'(g(x)).g'(x))color(white)(a/a)|)))color(green)" A"# #"------------------------------------------------------------"# here #f(g(x))=ln(x/(x-1))rArrf'(g(x))=1/(x/(x-1))=(x-1)/x# and #g(x)=x/(x-1)rArrg'(x) = "see below"# #"-----------------------------------------------------------"# To differentiate g(x) we require to use the #color(blue)"quotient rule"# If f(x)#=(g(x))/(h(x))" then " f'(x)=(h(x).g'(x)-g(x).h'(x))/(h(x)^2)color(green)"B"# #"-----------------------------------------------------------------"# here g(x) = x #rArrg'(x)=1# and h(x) = x-1 #rArrh'(x)=1# #"----------------------------------------------------------------"# Substitute these values into #color(green)" B"# #rArrf'(x)=((x-1).1-x.1)/(((x-1)^2))=(x-1-x)/(x-1)^2=(-1)/(x-1)^2# #"------------------------------------------------------------"# Now this is the value of g'(x) that we set out to obtain from above Finally , 'plug' these all back into #color(green)"A"# #rArrdy/dx=(x-1)/x xx(-1)/(x-1)^2# #=cancel((x-1))/x xx(-1)/(cancel((x-1)) (x-1))=(-1)/(x(x-1))#

Adam Adkins · Answer

undefined To differentiate $ y = \ln\left(\frac{x}{x-1}\right) $:

Use the chain rule and the properties of logarithms:

$$ \frac{dy}{dx} = \frac{1}{\frac{x}{x-1}} \cdot \frac{d}{dx}\left(\frac{x}{x-1}\right) $$

$$ \frac{dy}{dx} = \frac{1}{\frac{x}{x-1}} \cdot \left(\frac{(x-1) - x(1)}{(x-1)^2}\right) $$

$$ \frac{dy}{dx} = \frac{1}{\frac{x}{x-1}} \cdot \left(\frac{-1}{(x-1)^2}\right) $$

$$ \frac{dy}{dx} = \frac{-1}{x} $$