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Mathematics 18 Online
OpenStudy (sasogeek):

3rd derivative of f(x)=x

OpenStudy (sasogeek):

i know dy/dx=1, d2y/dx^2=0, but then what next?

OpenStudy (anonymous):

The derivative of zero is... get this... zero!

OpenStudy (sasogeek):

sigh, so that's it?

OpenStudy (anonymous):

http://www.wolframalpha.com/input/?i=derivative+of+zero There's like proof and stuff. Also, remember y''=0 is still a constant. Constant means no slope.

OpenStudy (sasogeek):

now i'm going to ask out teacher why he gave us that

OpenStudy (accessdenied):

Yep. If you graphed y=0, you could see that it is a horizontal line, slope 0

OpenStudy (anonymous):

You can prove it directly from the definition of a derivative. Let \(f(x)=0\) then \[f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{0-0}{h}=\lim_{h\to 0}0=0.\]

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