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

1 Answer
Jim H
May 5, 2016

Please see the explanation section below.

Explanation:

As with most questions involving the limit definition of derivative, we'll expand and simplify the difference quotient until we can evaluate the limit.
For this function we may need more than usual patience, determination and attention to detail.

#f(x) = (x^2-1)/(2x-3)#

The difference quotient:

#(f(x+h)-f(x))/h = (((x+h)^2-1)/(2(x+h)-3)- (x^2-1)/(2x-3))/h#

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

# = ((2x-3)((x+h)^2-1) - (x^2-1)(2(x+h)-3))/(h(2(x+h)-3)(2x-3))#

# = ((2x-3)(x^2+2xh+h^2-1) - (x^2-1)(2x+2h-3))/(h(2(x+h)-3)(2x-3))#

#(2x^3+4x^2h+2xh^2-2x-3x^2-6xh-3h^2+3-(2x^3+2x^2h-3x^2-2x-2h+3))/(h(2x+2h-3)(2x-3))#

# = (2x^2h+2xh^2-6xh-3h^2+2h)/(h(2x+2h-3)(2x-3))#

# =(cancel(h) (2x^2+2xh-6x-3h+2))/(cancel(h)(2x+2h-3)(2x-3))#

We can now evaluate the limit as #hrarr0#.

#f'(x) = lim_(hrarr0)(2x^2+2xh-6x-3h+2)/((2x+2h-3)(2x-3))#

# = (2x^2-6x+2)/((2x-3)(2x-3))#

# = (2x^2-6x+2)/(2x-3)^2#

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