anonymous
  • anonymous
let f(x)=cubed root of x If a does not =0 find f'(a) using the definition of a derivative.
Mathematics
chestercat
  • chestercat
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
definition, not the power rule right?
anonymous
  • anonymous
yeah, I am having some trouble because of the cubed root

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misty1212
  • misty1212
i bet i can show you the gimmick did you do it with the square root ever?
anonymous
  • anonymous
yeah I have
misty1212
  • 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
misty1212
  • 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. \[a^3-b^3=(a-b)( a^2+ab+b^2)\]
anonymous
  • anonymous
ohhh I was doing it as though it was a squared .-.
anonymous
  • anonymous
okay so how exactly will that look for this problem
misty1212
  • misty1212
\[\frac{\sqrt[3]{a+h}-\sqrt[3]{a}}{h}\] multiply top and bottom by \[\sqrt[3]{(a+h)^2}+\sqrt[3]{a}\sqrt[3]{a+h}+\sqrt[3]{a^2}\]
misty1212
  • misty1212
don't really multiply it out the numerator will be \[a+h-a=h\]
misty1212
  • misty1212
leave the denominator in factored form cancel the \(h\) top and bottom and then replace \(h\) by \(0\)
misty1212
  • misty1212
you are using this \[(a-b)(a^2+ab+b^2)=a^3-b^3\]with \[a=\sqrt[3]{a+h},b=\sqrt[3]{a}\]
anonymous
  • anonymous
I am still a a but confused..
anonymous
  • anonymous
*a bit
anonymous
  • anonymous

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