i need to use the product rule f(x)= (sgaure root of x)(sinx) and what I have is sqaure root of x (cosx) + (sinx)(1/2 x ^ -1/2) Is that right?? Doesn't seem like it as I am stuck.
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$\displaystyle \displaystyle (\sqrt{x}\sin x )' = \frac{1}{2\sqrt{x}}\sin x + \sqrt{x}\cos x$
Last edited by pickslides; Feb 13th 2011 at 01:45 PM. Reason: Bad Latex
i have no clue how you got that.....
Originally Posted by MZTakara i have no clue how you got that..... pickslides just cleaned up the algebra ... note that $\displaystyle \dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}$
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