How do you use the limit definition to find the derivative of #f(x)=x^3+1#?

Question

Charles Anctil · Accepted Answer

# f'(x) =3x^2 # By definition # f'(x) =lim_(hrarr0)(f(x+h)-f(x))/h # So, with # f(x)=x^3+1 # we have: # f'(x) =lim_(hrarr0) (((x+h)^3+1 ) - (x^3+1) ) / h# # :. f'(x) =lim_(hrarr0) (((x^3+3hx^2+3h^2x+h^3)+1 ) - x^3-1 ) / h# # :. f'(x) =lim_(hrarr0) (x^3+3hx^2+3h^2x+h^3+1 - x^3-1 ) / h# # :. f'(x) =lim_(hrarr0) ( 3hx^2+3h^2x+h^3 ) / h# # :. f'(x) =lim_(hrarr0) ( 3x^2+3hx+h^2 ) # # :. f'(x) =3x^2 #

Charles Arocha · Answer

undefined To find the derivative of $ f(x) = x^3 + 1 $ using the limit definition of a derivative, follow these steps:

1. Start with the definition of the derivative:

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

2. Substitute the function $ f(x) = x^3 + 1 $ into the definition:

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

3. Expand and simplify the numerator:

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

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

$$ f'(x) = \lim_{{h \to 0}} (3x^2 + 3xh + h^2) $$

4. Cancel out $ h $ in the numerator and denominator:

$$ f'(x) = 3x^2 + 3xh + h^2 $$

5. Take the limit as $ h $ approaches 0:

$$ f'(x) = 3x^2 + 0 + 0^2 $$

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

So, the derivative of $ f(x) = x^3 + 1 $ is $ f'(x) = 3x^2 $.