How do you find the derivative of (x^2-4)/(x-1)?

Mar 12, 2017

$\frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} - 4}{x - 1}\right) = 1 + \frac{3}{x - 1} ^ 2$

Explanation:

I would simplify the expression first, then use the power rule...

$\frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} - 4}{x - 1}\right) = \frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} - x + x - 1 - 3}{x - 1}\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} - 4}{x - 1}\right)} = \frac{d}{\mathrm{dx}} \left(\frac{\left(x + 1\right) \left(x - 1\right) - 3}{x - 1}\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} - 4}{x - 1}\right)} = \frac{d}{\mathrm{dx}} \left(\left(x + 1\right) - \frac{3}{x - 1}\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} - 4}{x - 1}\right)} = \frac{d}{\mathrm{dx}} \left(x + 1 - 3 {\left(x - 1\right)}^{- 1}\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} - 4}{x - 1}\right)} = 1 + 0 + 3 {\left(x - 1\right)}^{- 2}$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} - 4}{x - 1}\right)} = 1 + \frac{3}{x - 1} ^ 2$

$\textcolor{w h i t e}{}$
Note: I did use the chain rule quietly above too, when differentiating:

${\left(x - 1\right)}^{- 1}$

to get:

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