anonymous
  • anonymous
Can someone please help? In Paul's Online Math Notes, Applications of the Derivative, Optimization, example 2, how does he get the second derivative? How is C'(w)=120w-180w^-2=120w^3-800/w^2??? I get how to get the derivative, but not how the derivative is equal to the answer he gives?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
120w-180w^-2 is the same as 120w-180/w^2, right. So the process there is to get a common denominator, w^2, so it can all be written as one fraction.
anonymous
  • anonymous
I can see how it's re-written, but when I go to re-write it I want to write (120w-800)/w^2...where does 120w^3 come from?
anonymous
  • anonymous
You can't do that. See, if you split that up again you get something different right because you get 120w/w^2-800/w^2 which is not what you started with. You multiply 120w (think of it as 120w/1) by w^2/w^2 (which is just 1) so that it also has a denominator of w^2 and you can just add them.

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anonymous
  • anonymous
haha, yeah, I got it now...that's me forgetting my arithmetic...thanks!
anonymous
  • anonymous
It happens. Np

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