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

The definition of derivative is

With a little bit of patience, we can expand the numerator, and we'll see that many things will erase: the numerator only is equal to

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To find ( f'(x) ) using the definition of a derivative for ( \frac{x^2 + 1}{x - 2} ), we use the limit definition of a derivative:

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

First, we substitute ( f(x) = \frac{x^2 + 1}{x - 2} ) into the definition:

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

Next, we simplify the expression:

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

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

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

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

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

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

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

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

[ f'(x) = 0 ]

Therefore, ( f'(x) = 0 ).

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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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