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

Jan 26, 2016

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

#### Explanation:

$f \left(x\right) = \frac{x}{4 x - 2} = \frac{1}{2} \left(\frac{x}{2 x - 1}\right)$

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

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[\frac{1}{2} \left(\frac{x}{2 x - 1}\right)\right] = \frac{1}{2} \frac{d}{\mathrm{dx}} \left(\frac{x}{2 x - 1}\right) = \frac{1}{2} \frac{d}{\mathrm{dx}} i \left(x\right)$

Now $i \left(x\right) = \frac{h \left(x\right)}{g} \left(x\right)$

The Quotient Rule tells us:

given: $i \left(x\right) = \frac{h \left(x\right)}{g} \left(x\right)$

$i ' \left(x\right) = \frac{h ' \left(x\right) \cdot g \left(x\right) - g ' \left(x\right) \cdot h \left(x\right)}{g \left(x\right)} ^ 2$

$\therefore f ' \left(x\right) = \frac{1}{2} \left[\frac{1 \cdot \left(2 x - 1\right) - \left(2 - 0\right) \cdot x}{2 x - 1} ^ 2\right] =$

$= \frac{1}{2} \left[\frac{\textcolor{g r e e n}{\cancel{2 x}} - 1 \textcolor{g r e e n}{\cancel{- 2 x}}}{2 x - 1} ^ 2\right] =$

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