Question

Cat este [f'(x)]' daca f(x)=√x, am facut f'(x)=1/(2√x) si nu imi da mai departe​

Answer 1

Salut,

$(x^a)' = a\cdot x^{a-1},\ unde\ a\in\mathbb{R}.\\\\(\sqrt x)'=(x^{\frac{1}2})'=\dfrac{1}2\cdot x^{\frac{1}2-1}=\dfrac{1}2\cdot x^{-\frac{1}2}=\dfrac{1}{2\cdot\sqrt x}.\\\\\left(\dfrac{1}{2\cdot\sqrt x}\right)'=\dfrac{1}2\cdot(x^{-\frac{1}2})'=-\dfrac{1}4\cdot x^{-\frac{1}2-1}=-\dfrac{1}4\cdot x^{-\frac{3}2}=-\dfrac{1}{4\cdot\sqrt x^3}=-\dfrac{1}{4\cdot x\cdot\sqrt x}.$

Green eyes.

Answer 2

$f(x) = \sqrt x= x^{\dfrac{1}{2}} \\ \\ f'(x) = \dfrac{1}{2}x^\Big{\dfrac{1}{2}-1} =\dfrac{1}{2}x^{-\dfrac{1}{2}}\\ \\ f''(x) =-\dfrac{1}{4}x^\Big{-\dfrac{1}{2}-1} = -\dfrac{1}{4}x^{-\dfrac{3}{2}} = -\dfrac{{1}}{4\sqrt{x^3}}}$