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

1 Answer
Alberto P.
Nov 10, 2016

#2*(x^2-3x+1)/(2x-3)^2#

Explanation:

#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#

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