Answer
$\displaystyle \frac{dy}{dx}=\frac{3}{2}\sqrt{x}$
Work Step by Step
$(\sqrt{x}=^{1/2})$
and, for $\sqrt{x}$ to be defined , x can not be negative ,
so, if needed, $\sqrt{x^{2}}=|x|=x \qquad (*)$
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$f(x)=x,\ \ \quad g(x)=x^{1/2},\quad y=f(x)\cdot g(x)$
$f^{\prime}(x)=1\displaystyle \qquad g^{\prime}(x)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[f(x)\cdot g(x)]$= ... product rule ...
=$f^{\prime}(x)g(x)-f(x)g^{\prime}(x)$
=$ 1\displaystyle \cdot\sqrt{x}+x\cdot\frac{1}{2\sqrt{x}}$
$=\displaystyle \sqrt{x}+\frac{x}{2\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}}\quad $...see (*), $\sqrt{x}\sqrt{x}=x$
$=\displaystyle \sqrt{x}+\frac{x\sqrt{x}}{2x}$
$=\displaystyle \sqrt{x}+\frac{\sqrt{x}}{2}$
$=\displaystyle \frac{3}{2}\sqrt{x}$
