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

1. misty1212

HI!!

2. misty1212

definition, not the power rule right?

3. anonymous

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

4. misty1212

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

5. anonymous

yeah I have

6. misty1212

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

7. misty1212

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. $a^3-b^3=(a-b)( a^2+ab+b^2)$

8. anonymous

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

9. anonymous

okay so how exactly will that look for this problem

10. misty1212

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

11. misty1212

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

12. misty1212

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

13. misty1212

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

14. anonymous

I am still a a but confused..

15. anonymous

*a bit

16. anonymous

@misty1212