Using the limit definition, how do you find the derivative of  f(x) = (x^2-1) / (2x-3)?

Nov 10, 2016

$2 \cdot \frac{{x}^{2} - 3 x + 1}{2 x - 3} ^ 2$

Explanation:

f(x)=x/2+3/4+5/(4(2x-3)

$f ' \left({x}_{0}\right) = {\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)}{h} =$
$= {\lim}_{h \to 0} \left(\frac{\cancel{{x}_{0}} + \cancel{h} - \cancel{{x}_{0}}}{2 \cancel{h}} + \frac{5}{4} \cdot \frac{\frac{1}{2 {x}_{0} + 2 h - 3} - \frac{1}{2 {x}_{0} - 3}}{h}\right) =$
$= \frac{1}{2} + \frac{5}{4} {\lim}_{h \to 0} \frac{\cancel{2 {x}_{0}} \cancel{- 3} - \left(\cancel{2 {x}_{0}} + 2 h \cancel{- 3}\right)}{h \left(2 {x}_{0} - 2 h - 3\right) \left(2 {x}_{0} - 3\right)} =$
$= \frac{1}{2} + \frac{5}{4} {\lim}_{h \to 0} \frac{- 2 \cancel{h}}{\cancel{h} \left(2 {x}_{0} - 2 h - 3\right) \left(2 {x}_{0} - 3\right)} =$
$= \frac{1}{2} - \frac{5}{2} \cdot \frac{1}{2 {x}_{0} - 3} ^ 2 = \frac{\left(4 {x}_{0}^{2} + 9 - 12 {x}_{0}\right) - 5}{2 {\left(2 {x}_{0} - 3\right)}^{2}} =$
$= 2 \cdot \frac{{x}_{0}^{2} - 3 x + 1}{2 {x}_{0} - 3} ^ 2$