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

1 Answer
George C.
Nov 1, 2015

Long divide then use limit definition of derivative to find:

#f'(x) = 1/2 - 5/(2(2x-3)^2)#

Explanation:

Long divide:

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...to find...

#f(x) = (x^2-1)/(2x-3)=1/2x+3/4+5/(8x-12)#

Then using the limit definition of derivative:

#f'(x) = lim_(h->0) ((f(x+h) - f(x))/h)#

#=lim_(h->0) ((1/2(x+h)+3/4+5/(8(x+h)-12)) - (1/2x+3/4+5/(8x-12)))/h#

#=lim_(h->0) (1/2h+(5/(8(x+h)-12)-5/(8x-12)))/h#

#=1/2+lim_(h->0) ((5/(8(x+h)-12)-5/(8x-12))/h)#

#=1/2+lim_(h->0) ((5((8x-12)-(8(x+h)-12)))/(h(8(x+h)-12)(8x-12)))#

#=1/2+lim_(h->0) ((-40h)/(h(8(x+h)-12)(8x-12)))#

#=1/2+lim_(h->0) ((-40)/((8(x+h)-12)(8x-12)))#

#=1/2-40/((8x-12)^2)#

#=1/2-40/(16(2x-3)^2)#

#=1/2-5/(2(2x-3)^2)#

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