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How could one differentiate $x\sqrt{x}$?

I know $[\sqrt{x}]' = \dfrac{1}{2\sqrt{x}}$

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    Note the in the answers you have 2 different approaches: (1) $x^{\frac{2}{3}}$ and use the *power rule*. (2) $x \times \sqrt{x}$ and use the *product rule*.2012-10-25

2 Answers 2

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$\sqrt{x}=x^{1/2}$, so just differentiate $x^{3/2}$, i.e. $\frac{d}{dx}x\sqrt{x}=\frac{3}{2}\sqrt{x}$.

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Product rule: $(fg)'= f'g + fg'$

Take $f=x$ and $g=\sqrt x$

You know $f'=1$ so $ (fg)'(x)= \sqrt{x} + \frac{x}{2\sqrt{x}}. $