How do you differentiate #y=(1+lnx)/(x^2-lnx)# using the quotient rule?

Question

1931 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-28T13:47:02

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

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

#d/dx(p/q) = (qp'-pq')/q^2#

#d/dx(1+lnx)=1/x#

#d/dx(x^2-lnx) = 2x-1/x#

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

Expanding #dy/dx# and multiplying by #x/x# gives

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