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:

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

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

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