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

Answer 1
Use #f'(x) = lim_(h->0) (f(x + h) - f(x))/h#.
#f'(x) = lim_(h-> 0) (((x +h)^2 - 1)/(2(x + h) - 3) - (x^2 - 1)/(2x - 3))/h#
#f'(x) = lim_(h->0) (((x^2 + 2xh + h^2 - 1)(2x - 3))/((2x + 2h - 3)(2x -3)) - ((x^2 - 1)(2x + 2h - 3))/((2x + 2h - 3)(2x - 3)))/h#
#f'(x) = lim_(h->0) ((2x^3 + 4x^2h + 2h^2x - 2x - 3x^2 - 6xh -3h^2 + 3 - (2x^3 + 2x^2h - 3x^2 - 2x- 2h + 3))/((2x + 2h - 3)(2x -3)))/h#
#f'(x) = lim_(h->0) ((2x^3 + 4x^2h + 2h^2x - 2x - 3x^2 - 6xh - 3h^2 + 3 - 2x^3 - 2x^2h + 3x^2 + 2x + 2h - 3)/((2x + 2h - 3)(2x - 3)))/h#
#f'(x) = lim_(h->0) ((4x^2h + 2h^2x- 3h^2 - 6xh - 2x^2h+ 2h)/((2x + 2h - 3)(2x -3)))/h#
#f'(x) = lim_(h->0) ((2h^2x + 2x^2h - 3h^2 - 6xh + 2h)/((2x + 2h - 3)(2x - 3)))/h#
#f'(x) = lim_(h->0) (2h^2x + 2x^2h- 3h^2 - 6xh + 2h)/(h(2x + 2h - 3)(2x - 3))#
#f'(x)= lim_(h->0) (h(2hx + 2x^2 - 3h - 6x + 2))/(h(2x+ 2h -3)(2x - 3))#
#f'(x) = lim_(h->0) (2hx +2x^2- 3h - 6x +2)/((2x + 2h - 3)(2x -3)#
#f'(x) = (2(0) x + 2x^2 - 3(0) - 6x + 2)/((2x + 2(0) - 3)(2x - 3))#
#f'(x) = (2x^2 - 6x + 2)/(2x - 3)^2#
#f'(x) = (2(x^2 - 3x + 1))/(2x - 3)^2#

Verification using the quotient rule yields the same result.

Hopefully this helps!

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Answer 2

To find the derivative of (f(x) = \frac{x^2 - 1}{2x - 3}) using the definition of a derivative, you would first rewrite (f(x)) as a limit:

[ f'(x) = \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h} ]

Then, substitute (f(x)) into the limit expression:

[ f'(x) = \lim_{{h \to 0}} \frac{\frac{{(x + h)^2 - 1}}{{2(x + h) - 3}} - \frac{{x^2 - 1}}{{2x - 3}}}{h} ]

Simplify the expression:

[ f'(x) = \lim_{{h \to 0}} \frac{{(x^2 + 2hx + h^2 - 1) - (x^2 - 1)}}{h(2x - 3)} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{2hx + h^2}}{{h(2x - 3)}} ]

[ f'(x) = \lim_{{h \to 0}} \frac{{h(2x + h)}}{{h(2x - 3)}} ]

Cancel out (h) from the numerator and denominator:

[ f'(x) = \lim_{{h \to 0}} \frac{{2x + h}}{{2x - 3}} ]

Now, evaluate the limit as (h) approaches (0):

[ f'(x) = \frac{{2x}}{{2x - 3}} ]

So, the derivative of (f(x) = \frac{x^2 - 1}{2x - 3}) is (f'(x) = \frac{{2x}}{{2x - 3}}).

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Answer 3

To find the derivative of ( f(x) = \frac{x^2 - 1}{2x - 3} ), you can use the definition of a derivative, which states:

[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

Firstly, express ( f(x) ) as a fraction:

[ f(x) = \frac{x^2 - 1}{2x - 3} ]

Then, apply the limit definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{\left(\frac{(x+h)^2 - 1}{2(x+h) - 3}\right) - \left(\frac{x^2 - 1}{2x - 3}\right)}{h} ]

Next, simplify the expression:

[ f'(x) = \lim_{h \to 0} \frac{\left(\frac{x^2 + 2xh + h^2 - 1}{2x + 2h - 3}\right) - \left(\frac{x^2 - 1}{2x - 3}\right)}{h} ]

[ f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 - 1)(2x - 3) - (x^2 - 1)(2x + 2h - 3)}{h(2x + 2h - 3)(2x - 3)} ]

[ f'(x) = \lim_{h \to 0} \frac{2x^3 - 3x^2 + 2x^2h - 3xh + 2xh^2 - 3h - 2x^3 - 2x^2h + 3x^2 - 3x - 2h^2 + 3}{h(2x + 2h - 3)(2x - 3)} ]

[ f'(x) = \lim_{h \to 0} \frac{-3xh + 2xh^2 - 3h + 3}{h(2x + 2h - 3)(2x - 3)} ]

Finally, evaluate the limit as ( h ) approaches ( 0 ) to find the derivative.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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