# How do you find the derivative of quotient (1-3x)/(1+3x)?

Apr 12, 2015

The Quotient Rule for Derivatives says
$\frac{d \frac{g \left(x\right)}{h \left(x\right)}}{\mathrm{dx}} = \frac{g ' \left(x\right) \cdot h \left(x\right) - g \left(x\right) \cdot h ' \left(x\right)}{{h}^{2} \left(x\right)}$

So
$\frac{d \left(\frac{1 - 3 x}{1 + 3 x}\right)}{\mathrm{dx}}$

$= \frac{\left(- 3\right) \left(1 + 3 x\right) - \left(1 - 3 x\right) \left(3\right)}{{\left(1 + 3 x\right)}^{2}}$

$= \frac{- 6}{9 {x}^{2} + 6 x + 1}$

Apr 12, 2015

$\frac{- 6}{{\left(1 + 3 x\right)}^{2}}$

Detail

Let

$y = \frac{1 - 3 x}{1 + 3 x}$

Differentiating both sides with respect to 'x'

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{1 - 3 x}{1 + 3 x}\right)$

Using quotient rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 + 3 x\right) \frac{d}{\mathrm{dx}} \left(1 - 3 x\right) - \left(1 - 3 x\right) \frac{d}{\mathrm{dx}} \left(1 + 3 x\right)}{{\left(1 + 3 x\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 + 3 x\right) \left(0 - 3\right) - \left(1 - 3 x\right) \left(0 + 3\right)}{{\left(1 + 3 x\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(- 3\right) \left(1 + 3 x\right) - \left(3\right) \left(1 - 3 x\right)}{{\left(1 + 3 x\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 \left(1\right) + \left(- 3\right) \left(3 x\right) - 3 \left(1\right) - 3 \left(- 3 x\right)}{{\left(1 + 3 x\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 - 9 x - 3 + 9 x}{{\left(1 + 3 x\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 - 3 + \cancel{9 x} - \cancel{9 x}}{{\left(1 + 3 x\right)}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 6}{{\left(1 + 3 x\right)}^{2}}$