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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.
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.

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misty1212
 one year ago
Best ResponseYou've already chosen the best response.1definition, not the power rule right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, I am having some trouble because of the cubed root

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1i bet i can show you the gimmick did you do it with the square root ever?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1ok the idea with the square root is to multiply by the conjugate, because \[a^2b^2=(a+b)(ab)\] so the radical goes away

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1but 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. \[a^3b^3=(ab)( a^2+ab+b^2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohhh I was doing it as though it was a squared ..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so how exactly will that look for this problem

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{\sqrt[3]{a+h}\sqrt[3]{a}}{h}\] multiply top and bottom by \[\sqrt[3]{(a+h)^2}+\sqrt[3]{a}\sqrt[3]{a+h}+\sqrt[3]{a^2}\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1don't really multiply it out the numerator will be \[a+ha=h\]

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1leave the denominator in factored form cancel the \(h\) top and bottom and then replace \(h\) by \(0\)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1you are using this \[(ab)(a^2+ab+b^2)=a^3b^3\]with \[a=\sqrt[3]{a+h},b=\sqrt[3]{a}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am still a a but confused..
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