y= √x + ln(x)
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Originally Posted by abii y= √x + ln(x) $\displaystyle (\ln x)' = \frac{1}{x}$ and $\displaystyle (\sqrt{x})' = \frac{-1}{2\sqrt{x}}$
Originally Posted by pickslides $\displaystyle (\ln x)' = \frac{1}{x}$ and $\displaystyle (\sqrt{x})' = \frac{-1}{2\sqrt{x}}$ ... $\displaystyle \textcolor{red}{\frac{+1}{2\sqrt{x}}}$ , typo? ...
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