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

Question

Christopher Allers · Accepted Answer

#2*(x^2-3x+1)/(2x-3)^2# #f(x)=x/2+3/4+5/(4(2x-3)# #f'(x_0)=lim_(h->0)(f(x_0+h)-f(x_0))/h=# #=lim_(h->0)((cancel(x_0)+cancel(h)-cancel(x_0))/(2cancel(h))+5/4*(1/(2x_0+2h-3)-1/(2x_0-3))/h)=# #=1/2+5/4lim_(h->0)(cancel(2x_0)cancel(-3)-(cancel(2x_0)+2hcancel(-3)))/(h(2x_0-2h-3)(2x_0-3))=# #=1/2+5/4lim_(h->0)(-2cancel(h))/(cancel(h)(2x_0-2h-3)(2x_0-3))=# #=1/2-5/2*1/(2x_0-3)^2=((4x_0^2+9-12x_0)-5)/(2(2x_0-3)^2)=# #=2*(x_0^2-3x+1)/(2x_0-3)^2#

Luke Alvarez · Answer

undefined To find the derivative of the function $ f(x) = \frac{x^2 - 1}{2x - 3} $ using the limit definition, you would follow these steps:

1. Start with the definition of the derivative: $ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $.

2. Substitute the given function into the definition: $ f'(x) = \lim_{h \to 0} \frac{\frac{(x + h)^2 - 1}{2(x + h) - 3} - \frac{x^2 - 1}{2x - 3}}{h} $.

3. Simplify the expression inside the limit by finding common denominators and combining terms.

4. Once the expression is simplified, evaluate the limit as $ h $ approaches 0 to find the derivative.