How do you find derivative of #y=(-2lnx)/(3x-1)#?

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Answer 1 · 2015-03-06T22:21:01

How: Use the quotient rule and the fact that #d/(dx)(lnx)=1/x#.

The quotient rule for derivatives says that: #d/(dx)(N/D)=(N'D-ND')/D^2#

So, #y=(-2lnx)/(3x-1)# has derivative:

#y'=(-2(1/x)(3x-1)-(-2lnx)(3))/(3x-1)^2# (we're finished with the calculus).

Use algebra to re-write:

#y'=(((-2(3x-1))/x)+6lnx)/(3x-1)^2=((2-6x+6xlnx)/x)/(3x-1)^2#

#y'=(2-6x+6xlnx)/(x(3x-1)^2)#