# How do you differentiate (4x^2 -6x + 9 )/ (2x-3) using the quotient rule?

Sep 20, 2016

Derivative of $\frac{4 {x}^{2} - 6 x + 9}{2 x - 3}$ is $\frac{8 x \left(x - 3\right)}{2 x - 3} ^ 2$

#### Explanation:

Quotient rule states if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) - \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)}{h \left(x\right)} ^ 2$

Hence as $f \left(x\right) = \frac{4 {x}^{2} - 6 x + 9}{2 x - 3}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\left(4 \times 2 x - 6\right) \times \left(2 x - 3\right) - 2 \times \left(4 {x}^{2} - 6 x + 9\right)}{2 x - 3} ^ 2$

= $\frac{\left(8 x - 6\right) \left(2 x - 3\right) - 8 {x}^{2} + 12 x - 18}{2 x - 3} ^ 2$

= $\frac{8 x \times 2 x - 3 \times 8 x - 6 \times 2 x - 6 \times \left(- 3\right) - 8 {x}^{2} + 12 x - 18}{2 x - 3} ^ 2$

= $\frac{16 {x}^{2} - 24 x - 12 x + 18 - 8 {x}^{2} + 12 x - 18}{2 x - 3} ^ 2$

= $\frac{8 {x}^{2} - 24 x}{2 x - 3} ^ 2$

= $\frac{8 x \left(x - 3\right)}{2 x - 3} ^ 2$