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

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

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Do you know the product rule for derivatives? – anonymous Oct 25 '12 at 17:26
Or you can write $x\sqrt x=x^{\frac 32}$ and use the power law. – Ross Millikan Oct 25 '12 at 17:27
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. – user2468 Oct 25 '12 at 17:36
up vote 5 down vote accepted

$\sqrt{x}=x^{1/2}$, 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 $$ (fg)'(x)= \sqrt{x} + \frac{x}{2\sqrt{x}}. $$

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