let f(x)=cubed root of x
If a does not =0 find f'(a) using the definition of a derivative.

Question

misty1212 · Answer

HI!!

misty1212 · Answer

definition, not the power rule right?

anonymous · Answer

yeah, I am having some trouble because of the cubed root

misty1212 · Answer

i bet 
i can show you the gimmick
did you do it with the square root ever?

anonymous · Answer

yeah I have

misty1212 · Answer

ok the idea with the square root is to multiply by the conjugate, because $$a^2-b^2=(a+b)(a-b)$$ so the radical goes away

misty1212 · Answer

but for the cubed root you have to use the difference of two cubes, not the difference of two squares 
that will get rid of the cubed root
i.e. $$a^3-b^3=(a-b)(
a^2+ab+b^2)$$

anonymous · Answer

ohhh I was doing it as though it was a squared .-.

anonymous · Answer

okay so how exactly will that look for this problem

misty1212 · Answer

$$\frac{\sqrt[3]{a+h}-\sqrt[3]{a}}{h}$$ multiply top and bottom by $$\sqrt[3]{(a+h)^2}+\sqrt[3]{a}\sqrt[3]{a+h}+\sqrt[3]{a^2}$$

misty1212 · Answer

don't really multiply it out
the numerator will be $$a+h-a=h$$

misty1212 · Answer

leave the denominator in factored form
cancel the $h$ top and bottom 
 and then replace $h$ by $0$

misty1212 · Answer

you are using this $$(a-b)(a^2+ab+b^2)=a^3-b^3$$with $$a=\sqrt[3]{a+h},b=\sqrt[3]{a}$$

anonymous · Answer

I am still a a but confused..

anonymous · Answer

*a bit

anonymous · Answer

@misty1212