Derivative of f: Finding the Derivative of \sqrt{x}(2x-7)

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The discussion focuses on finding the derivative of the function f(x) = √x(2x - 7) using the product rule. The initial derivative expression provided is correct: (1/2)x^(-1/2)(2x - 7) + √x(2). Participants clarify the application of the product rule and express some confusion regarding fractional exponents and radicals. There is a suggestion to factor the term (1/2)x^(-1/2) into (2x - 7) for simplification. Overall, the conversation revolves around confirming the correct application of calculus rules and addressing misunderstandings related to derivatives.
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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.
 
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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)
 
justinbaker said:
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.
 
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}) into the (2x+7) ?
 
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