# Thread: limit definition of the derivative...final exam review!

1. ## limit definition of the derivative...final exam review!

i cant quite remember how to do these,

Use the limit definition of derivative to calculate
f'(x) where f (x) = 1/

x^3

i want to make sure i understand all the practice examples thoroughly for my exam. i have some previous questions i still need help with, so help on those or this is greatly appreciated!

2. Originally Posted by twostep08
i cant quite remember how to do these,

Use the limit definition of derivative to calculate
f'(x) where f (x) = 1/

x^3

i want to make sure i understand all the practice examples thoroughly for my exam. i have some previous questions i still need help with, so help on those or this is greatly appreciated!
I assume that you want to use the definition

$\displaystyle f'(x)=\lim_{h \to 0}\left(\dfrac{f(x+h)-f(x)}{h}\right)$

If so:

$\displaystyle f'(x)=\lim_{h \to 0}\left(\dfrac{\frac1{(x+h)^3}-\frac1{x^3}}{h}\right) = \lim_{h \to 0}\left(\dfrac{x^3-(x+h)^3}{hx^3(x+h)^3}\right)$

Expand the terms in the numerator and collect like terms:

$\displaystyle f'(x)= \lim_{h \to 0}\left(\dfrac{-3hx^2-3h^2x-h^3}{hx^3(x+h)^3}\right)$

Factor out h in the numerator:

$\displaystyle f'(x)= \lim_{h \to 0}\left(\dfrac{h(-3x^2-3hx-h^2)}{hx^3(x+h)^3}\right)$

Cancel h.

Now h is approaching zero: Then the numerator becomes $\displaystyle -3x^2$ and the denominator becomes $\displaystyle x^6$

Therefore

$\displaystyle f'(x)=\lim_{h \to 0}\left(\dfrac{\frac1{(x+h)^3}-\frac1{x^3}}{h}\right) = \dfrac{-3x^2}{x^6} = -\dfrac3{x^4}$

3. thats what i initially thought of doing.. is there anything else the question could mean becuase ive only heard that referred to as "the formal definition of the derivitave" i believe

i certainly hope its the way you showed , because i understand that, but do you think its possible that the ? refers to another method?