How do you differentiate #y=ln((x-1)/(x^2+1))#?

Question

10855 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-27T11:39:47

#dy/dx=(-x^2+2x+1)/((x^2+1)(x-1))#

#y=ln((x-1)/(x^2+1))#

#y=ln(x-1)-ln(x^2+1)#

Use quotient rule of logarithms

Now differentiate

#dy/dx=1/(x-1)-1/(x^2+1) * d/dx(x^2+1) #Use chain rule

#dy/dx=1/(x-1)-1/(x^2+1)*2x#

# dy/dx=1/(x-1)-(2x)/(x^2+1)# Take the lcd as ((x-1) (x^2+1)

#dy/dx=((x^2+1)/((x^2+1)(x-1)) )-((2x)(x-1))/((x^2+1)(x-1)) )#
#dy/dx=(x^2+1-2x^2+2x)/((x^2+1)(x-1)#
#dy/dx=(-x^2+2x+1)/((x^2+1)(x-1))#