Question

derivada até segunda ordem de raiz de x.

Answer 1

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

Answer 2

$f_{(x)}=\sqrt{x}=x^{\frac{1}{2}}\\\\ Derivada\ primeira:\\\\ f'_{(x)}=\frac{1}{2}x^{\frac{1}{2}-1}\\\\ f'_{(x)}=\frac{1}{2}x^{-\frac{1}{2}}\\\\ Derivada\ segunda:\\\\ f''_{(x)}=-\frac{1}{2}.\frac{1}{2}x^{-\frac{1}{2}-1}\\\\ f''_{(x)}=-\frac{1}{4}x^{-\frac{3}{2}}\\\\ f''_{(x)}=-\frac{1}{4}\frac{1}{x^{\frac{3}{2}}}\\\\ f''_{(x)}=-\frac{1}{4}\frac{1}{\sqrt[2]{x^3}}\\\\ f''_{(x)}=-\frac{1}{4\sqrt[2]{x^3}}\\\\ f''_{(x)}=-\frac{1}{4\sqrt[2]{x^2 \times x}}\\\\ \boxed{f''_{(x)}=-\frac{1}{4x\sqrt[2]{x}}}$


$Racionalizando:\\\\ f''_{(x)}=-\dfrac{1}{4x\sqrt{x}}\times \dfrac{\sqrt{x}}{\sqrt{x}}\\\\\\ f''_{(x)}=-\dfrac{\sqrt{x}}{4x^2}$


Espero ter ajudado.
Bons estudos!