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

Question

Elijah Abram · Accepted Answer

When asked to differentiate using the limit definition, use the formula #f'(x) = lim_(h ->0) (f(x + h) - f(x))/h#.

Let the function be #f(x)#: #f'(x) = lim_(h ->0) (f(x + h) - f(x))/h# #f'(x) = lim_(h->0) (1/((x + h)^2 - 1) - 1/(x^2 - 1))/h# #f'(x) = lim_(h->0) (1/(x^2 + 2xh + h^2 - 1) - 1/(x^2 - 1))/h# #f'(x) = lim_(h->0) ((x^2 - 1 - (x^2 + 2xh + h^2 - 1))/((x^2 - 1)(x^2 + 2xh + h^2 - 1)))/h# #f'(x) = lim_(h->0) ((x^2 - 1-x^2 - 2xh-h^2+ 1)/((x^2 - 1)(x^2+ 2xh + h^2 - 1)))/h# #f'(x) = lim_(h->0) (-2xh + h^2)/(h(x^4 + 2x^3h + x^2h^2 - x^2 - x^2-2xh-h^2 + 1))# #f'(x) = lim_(h->0) (cancel(h)(-2x + h))/(cancel(h)(x^4 + 2x^3h + x^2h^2+ 2xh + h^2 + 1 - 2x^2)# Substituting: #f'(x) = (-2x + 0)/(x^4 + 2x^3(0) + x^2(0)^2 + 2x(0) + 0^2 + 1 - 2x^2)# #f'(x) = -(2x)/(x^4 - 2x^2+ 1)# Hence, the derivative of #f(x) = 1/(x^2 - 1)# is #f'(x) = -(2x)/(x^4 - 2x^2 + 1)#. All that we have left to do is plug in #x = 3# into our derivative: #f'(3) = -(2 xx 3)/(3^4 - 2(3)^2 + 1)# #f'(3) = -6/(81 - 18 + 1)# #f'(3) = -6/(64)# #f'(3) = -3/32# In summary, when #f'(3)# is evaluated inside the derivative of #f(x) = 1/(x^2 - 1)#, a result of #-3/32# is obtained. Hopefully this helps!

Charles Anctil · Answer

undefined To find $ f'(3) $ using the limit definition, we start with the function $ f(x) = \frac{1}{x^2 - 1} $ and use the definition of the derivative:

$$ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} $$

In this case, $ a = 3 $, so we have:

$$ f'(3) = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h} $$

Now, let's plug in the function $ f(x) = \frac{1}{x^2 - 1} $ and $ a = 3 $ into the limit definition:

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

Simplify the expression inside the limit:

$$ f'(3) = \lim_{h \to 0} \frac{\frac{1}{9 + 6h + h^2 - 1} - \frac{1}{8}}{h} $$

$$ f'(3) = \lim_{h \to 0} \frac{\frac{1}{8 + 6h + h^2} - \frac{1}{8}}{h} $$

Combine the fractions under the same denominator:

$$ f'(3) = \lim_{h \to 0} \frac{\frac{8 - (8 + 6h + h^2)}{8(8 + 6h + h^2)}}{h} $$

$$ f'(3) = \lim_{h \to 0} \frac{\frac{-6h - h^2}{8(8 + 6h + h^2)}}{h} $$

Now simplify and factor out $ h $ from the numerator:

$$ f'(3) = \lim_{h \to 0} \frac{-6 - h}{8(8 + 6h + h^2)} $$

$$ f'(3) = \frac{-6}{8(8)} $$

$$ f'(3) = \frac{-6}{64} $$

$$ f'(3) = -\frac{3}{32} $$

So, $ f'(3) = -\frac{3}{32} $ using the limit definition.