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anonymous
 5 years ago
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)=120w180w^2=120w^3800/w^2??? I get how to get the derivative, but not how the derivative is equal to the answer he gives?
anonymous
 5 years ago
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)=120w180w^2=120w^3800/w^2??? I get how to get the derivative, but not how the derivative is equal to the answer he gives?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0120w180w^2 is the same as 120w180/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
 5 years ago
Best ResponseYou've already chosen the best response.0I can see how it's rewritten, but when I go to rewrite it I want to write (120w800)/w^2...where does 120w^3 come from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't do that. See, if you split that up again you get something different right because you get 120w/w^2800/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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, yeah, I got it now...that's me forgetting my arithmetic...thanks!
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