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Calculus1 15 Online
OpenStudy (anonymous):

Find f ' (t) for F(t) = square root of ( t^2 + 9 )

OpenStudy (jbo11):

have you learned the quick way to do derivatives?

OpenStudy (jbo11):

or do you have to use (x+h)

OpenStudy (anonymous):

Rewrite F9t) as (t^2 + 9)^(1/2) Use chain rule, you get F ' (t) = (1/2)(t^2 + 9)^(-1/2) (2t) = t/sqrt(t^2+9)

OpenStudy (anonymous):

@jbo11 Not sure I know the ( x+h), so i guess i know the fast way. Maybe...idk

OpenStudy (anonymous):

So perhaps you ahvent learned the Chain Rule. You may have found the derivative via definition using limits; also very important.

OpenStudy (jbo11):

ok well for the fast way, Easyaspi314 did it right. the (x+h) is the long way

OpenStudy (anonymous):

I actually do know the chain rule, sometimes I have trouble with it though

OpenStudy (jbo11):

you can think of the chain rule as getting rid of the ( )

OpenStudy (jbo11):

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