How could one differentiate $x\sqrt{x}$?
I know $[\sqrt{x}]' = \dfrac{1}{2\sqrt{x}}$
How could one differentiate $x\sqrt{x}$?
I know $[\sqrt{x}]' = \dfrac{1}{2\sqrt{x}}$
$\sqrt{x}=x^{1/2}$, so just differentiate $x^{3/2}$, i.e. $\frac{d}{dx}x\sqrt{x}=\frac{3}{2}\sqrt{x}$.
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}}. $