How do you differentiate $y=x/(3x-1)$ ?

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Answer 1 · 2015-03-09T01:10:39

I would use the Quotient Rule to get:
$y'=(1*(3x-1)-x*3)/(3x-1)^2=$
$=(3x-1-3x)/(3x-1)^2=$
$=-1/(3x-1)^2$

The Quotient Rule tells you that;
given the function $y=f(x)/g(x)$
The derivative is:
$y'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$

How do you differentiate y=x/(3x-1)y=x/(3x-1)?