help w/ lim {X->0} [(x + delta x)^3 - x^3] / delta x

[ayen]

Ok so i'm new to this but i really need help. I realize that this may be easy to some people but i've had a long day and i'm drawing a blank. Any help i get would be greatly appreciated. Thanks in advance.

lim {X->0} [(x + delta x)^3 - x^3] / delta x

Normally i would attempt the problem but like i said i can't think of the first thing to do. I'm stuck. Please help if you can.

 

[soroban]

Re: help please

Hello, ayen!

I must I assume you've done one of these before.

Definition of the derivative: $\displaystyle \,f'(x) \:=\:\lim_{\Delta x\to0}\,\frac{f(x\,+\,\Delta x)\,-\,f(x)}{\Delta x}$

$\displaystyle \lim_{\Delta x\to0}\,\frac{(x\,+\,\Delta x)^3\,-\,x^3}{\Delta x}$

Make four steps out of it . . .

[1] Expand the cube: $\displaystyle \,(x\,+\,\Delta x)^3\;=\;x^3\,+\,3x^2)\Delta x) \,+\,3x(\Delta x)^2\,+\,(\Delta x)^3$

[2] Subtract $\displaystyle x^3:\;\;\left[x^3\,+\,3x^2(\Delta x) + 3x(\Delta x)^2\,+\,(\Delta x)^3\right]\,-\,x^3\;=\;3x^2(\Delta x)\,+\,3x(\Delta x)^2 \,+\,(\Delta x)^3$

[3] Divide by $\displaystyle \Delta x:\;$\(\displaystyle \;\frac{3x^2(\Delta x)\,+\,3x(\Delta x)^2\,+\,(\Delta x)^3}{\Delta x}\:=\:\frac{\Delta x\left[3x^2\,+\,3x(\Delta x)\,+\,(\Delta x)^2)}{\Delta x} \:=\:3x^2\,+\,3x(\Delta x)\,+\,(\Delta x)^2\)

[4] Take the limit: $\displaystyle \:\lim_{\Delta x\to0}\,\left[3x^2\,+\,3x(\Delta x)\,+\,(\Delta x)^2\right] \;=\;3x^2$

 

[ayen]

Thanks so much! You really helped me!