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Dantes
Sep20-04, 10:42 PM
Find the derivative of f when :

\sqrt{x}(2x-7)

When put into product rule form (following (f '*g) + (f*g') )

(\frac{1}{2}x^\frac{-1}{2})\ast(2x+7) + \sqrt{x}(2)

Just started today making sure I am on the right track.

justinbaker
Sep20-04, 10:50 PM
i am pretty sure this is actually the answer when in product rule

(\frac{1}{2}x^\frac{-1}{2})\ast(2x+7) + \sqrt{x}(2)

Dantes
Sep20-04, 10:52 PM
i am pretty sure this is actually the answer when in product rule

(\frac{1}{2}x^\frac{-1}{2})\ast(2x+7) + \sqrt{x}(2)

Ahh yes...My fault doubled it twice.. thanks.

Dantes
Sep20-04, 11:25 PM
hmm my fundamentals are screwed up, I forgot what happens here for the most part. with the fraction exponents and the radicals.

Factor the (\frac{1}{2}x^\frac{-1}{2}) in to the (2x+7) ?