How do you differentiate $f(x)= x/(4x-2)$ using the quotient rule?

Question

1662 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-26T13:02:40

$f'(x)=-1/(2(2x-1)^2$

$f(x)=x/(4x-2)=1/2(x/(2x-1))$

We could taking 2 at the denominator to make easy the computation

$f'(x)=d/(dx)[1/2(x/(2x-1))]=1/2d/(dx)(x/(2x-1))=1/2d/(dx)i(x)$

Now $i(x)=(h(x))/g(x)$

The Quotient Rule tells us:

given: $i(x)=(h(x))/g(x)$

$i'(x)=(h'(x)*g(x)-g'(x)*h(x))/(g(x))^2$

$:.f'(x)=1/2[(1*(2x-1)-(2-0)*x)/(2x-1)^2]=$

$=1/2[(color(green)cancel(2x)-1color(green)cancel(-2x))/(2x-1)^2]=$

$=1/2*(-1/(2x-1)^2)=-1/(2(2x-1)^2$

How do you differentiate f(x)= x/(4x-2)f(x)= x/(4x-2) using the quotient rule?