Differentiation quotient rule question, am i wrong or is my book wrong?

[sponsoredwalk]

Differentiate with respect to x; (using the quotient rule)
3/2x-1 (3 over 2x minus 1)
dy/dx = (2x-1)(0) - (3)(2) / (2x-1)^2
dy/dx = -6/(2x-1)^2

but my book gives -2/(2x-1)^2

now,
y = u/v and i take
u = 3 and
v = 2x-1.

dy/dx = v(du/dx) - u(dv/dx) / (v)^2

hmm...

 

[Elucidus]

Assuming the question is to find

\frac{d}{dx} \; \frac{3}{2x-1}

then your answer is correct. Whoever wrote the solutions probably forgot all about the 3.

--Elucidus