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

3 Answers
Jul 22, 2017

I got: #f'(x)=(2x^2-6x+2)/(2x-3)^2#

Explanation:

Have a look:

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Jul 22, 2017

Well it gets a bit messy.

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Jul 23, 2017

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

Explanation:

The definition of the derivative of #y=f(x)# is

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

So with # f(x) = (x^2-1)/(2x-3) # then;

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

And so the derivative of # y=f(x) # is given by:

# f'(t) = lim_(h rarr 0) ( ((x^2+2hx+h^2-1)/(2x+2h-3)) - ((x^2-1)/(2x-3)) ) / h #

# " " = lim_(h rarr 0) 1/h( (x^2+2hx+h^2-1)/(2x+2h-3) - (x^2-1)/(2x-3) )#

# " " = lim_(h rarr 0) ( (x^2+2hx+h^2-1)(2x-3) - (x^2-1)(2x+2h-3) ) / (h(2x+2h-3)(2x-3) )#

# " " = lim_(h rarr 0) ( (2x^3+4hx^2+2h^2x-2x-3x^2-6hx-3h^2+3) - (2x^3+2hx^2-3x^2-2x-2h+3) ) / (h(2x+2h-3)(2x-3) )#

# " " = lim_(h rarr 0) (2x^3+4hx^2+2h^2x-2x-3x^2-6hx-3h^2+3 -2x^3-2hx^2+3x^2+2x+2h+3) / (h(2x+2h-3)(2x-3) )#

# " " = lim_(h rarr 0) (2hx^2+2h^2x-6hx-3h^2+2h) / (h(2x+2h-3)(2x-3) )#

# " " = lim_(h rarr 0) (2x^2+2hx-6x-3h+2) / ((2x+2h-3)(2x-3) )#

# " " = (2x^2+0-6x-0+2) / ((2x+0-3)(2x-3) )#

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

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

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