How do you use the definition of a derivative to find f' given #f(x)=x^3# at x=2?

Question

Emma Aldridge · Accepted Answer

#f'(2) = 12#

Use the formula #f'(x) = lim_(h-> 0) (f(x + h) - f(x))/h#.

#f'(x) = lim_(h-> 0) ((x+ h)^3 - x^3)/h#

Expand the binomial either by tedious algebra or by the Binomial Theorem/Pascal's Triangle.

#(x + h)^3 = (x + h)(x + h)(x + h) = (x^2 + 2xh + h^2)(x + h) = x^3 + 2x^2h + h^2x + hx^2 + 2xh^2 + h^3#

#f'(x) = lim_(h->0) (x^3 + 2x^2h + h^2x + hx^2 + 2xh^2 + h^3 - x^3)/h#

#f'(x) = lim_(h->0) (h(2x^2 + hx + x^2 + 2xh + h^2))/h#

#f'(x) = lim_(h-> 0) 2x^2 + hx + x^2 + 2xh + h^2#

#f'(x) = 2x^2 + 0(x) + x^2 + 2x(0) + 0^2#

#f'(x) = 3x^2#

A check using the power rule yields similar results.

We now simply evaluate the given value within the derivative.

#f'(2) = 3(2)^2 = 12#

Hopefully this helps!

Adam Adkins · Answer

undefined To find $ f'(x) $ given $ f(x) = x^3 $ at $ x = 2 $ using the definition of a derivative:

1. Use the definition of the derivative: $ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $.
2. Substitute $ f(x) = x^3 $ into the formula: $ f'(x) = \lim_{h \to 0} \frac{(x + h)^3 - x^3}{h} $.
3. Expand $ (x + h)^3 $ using binomial expansion: $ (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 $.
4. Substitute the expanded expression into the formula: $ f'(x) = \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h} $.
5. Simplify the expression: $ f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h} $.
6. Cancel out common terms: $ f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2) $.
7. Plug in $ x = 2 $: $ f'(2) = 3(2)^2 + 3(2)(0) + 0^2 $.
8. Evaluate: $ f'(2) = 12 $.

Therefore, $ f'(2) = 12 $.