# How do you find the derivative of x=((s-1)(s-2))/(s-3)?

May 25, 2017

$\frac{\mathrm{dx}}{\mathrm{ds}} = \frac{{s}^{2} - 6 s + 7}{s - 3} ^ 2$

#### Explanation:

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

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

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

Hence $\frac{\mathrm{dx}}{\mathrm{ds}} = \frac{\left(2 s - 3\right) \left(s - 3\right) - \left({s}^{2} - 3 s + 2\right) \times 1}{s - 3} ^ 2$

= $\frac{2 {s}^{2} - 9 s + 9 - {s}^{2} + 3 s - 2}{s - 3} ^ 2$

= $\frac{{s}^{2} - 6 s + 7}{s - 3} ^ 2$