Salut,
[tex](x^a)' = a\cdot x^{a-1},\ unde\ a\in\mathbb{R}.\\\\(\sqrt x)'=(x^{\frac{1}2})'=\dfrac{1}2\cdot x^{\frac{1}2-1}=\dfrac{1}2\cdot x^{-\frac{1}2}=\dfrac{1}{2\cdot\sqrt x}.\\\\\left(\dfrac{1}{2\cdot\sqrt x}\right)'=\dfrac{1}2\cdot(x^{-\frac{1}2})'=-\dfrac{1}4\cdot x^{-\frac{1}2-1}=-\dfrac{1}4\cdot x^{-\frac{3}2}=-\dfrac{1}{4\cdot\sqrt x^3}=-\dfrac{1}{4\cdot x\cdot\sqrt x}.[/tex]
Green eyes.
