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

Question

Emilia Aguilar · Accepted Answer

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

Have a look:

Charles Arocha · Answer

Well it gets a bit messy.

Ezekiel Alexander · Answer

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

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#

Paisley Johnson · Answer

undefined To find $ f'(x) $ using the limit definition given $ f(x) = \frac{x^2 - 1}{2x - 3} $:

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

2. Substitute $ f(x) $ into the expression: $ 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 by finding a common denominator and combining fractions.

4. After simplification, evaluate the limit as $ h $ approaches 0. This will give the derivative $ f'(x) $.