Question

dada a função y raiz de x determine o valor de f"(4).

Answer 1

Seja f=√x=x^(1/2), aplicando a regra do expoente, obtemos:
f'=(1/2)x^(1/2-1)=(1/2)x^(-1/2)
e de forma análoga,
f''=-(1/4)x^(-3/2).

Answer 2

Pelo que entendi, f(x) = y√x. Assim,

f' = $(y \sqrt{x})' = y. \frac{d}{dx}\sqrt{x} = y. \frac{1}{2\sqrt{x}} = \frac{y}{2\sqrt{x}}$


f'' = 

$\frac{y}{2} . \frac{1}{\sqrt{x}} = \frac{y}{2} . \frac{d}{dx} x^{\frac{-1}{2}}  \\ = \frac{y}{2}.(\frac{-1}{2}x^{2} ) = \frac{-y}{4}.\frac{1}{x^{\frac{3}{2}}} \\ = -\frac{y}{4\sqrt{x^{3}}}$


f''(4) = $-\frac{y}{4\sqrt{4^{3}}} = -\frac{y}{4\sqrt{64}} = -\frac{y}{32}$