Question

Derivada de f(x)=Raiz de x ao cubo -2 sobre/ dividido por sen(x)

Answer 1

$f(x) = \frac{ \sqrt{x^3 - 2}}{sen(x)}$

Vamos usar a regra do quociente, dizendo que $g(x) = \sqrt{x^3 - 2}$ e $h(x) = sen(x)$

$f'(x) = \frac{g'(x) * h(x) - g(x) * h'(x)}{h(x)^2}$

$g'(x) = (x^3 - 2)^{\frac{1}{2} } \\u = x^3 - 2\\g'(x) = \frac{1}{2} * (x^3 - 2)^{\frac{1}{2} - 1} * \frac{du}{dx} \\g'(x) = \frac{1}{2} * (x^3 - 2)^{-\frac{1}{2}} * (3x^2) = \frac{3x^2}{2\sqrt{x^3 - 2}}$

$h'(x) = cos(x)$

$\frac{[\frac{3x^2}{2\sqrt{x^3-2} } ] * sen(x) - [\sqrt{x^3-2} * cos(x)]}{sen(x)^2} = \frac{\frac{3x^2 * sen(x)}{2\sqrt{x^3-2} } - cos(x) * \sqrt{x^3-2}}{sen(x)^2} \\\frac{\frac{3x^2 * sen(x) - 2cos(x)(x^3 - 2)}{2\sqrt{x^3-2}}}{sen(x)^2} = \frac{3x^2sen(x) - 2cos(x)(x^3-2)}{2sen(x)^2\sqrt{x^3-2} }$