Question

Найдите производную функции y = 3/x^3 + x^3/3 - 6 корень x

Answer 1

$\displaystyle (\frac{3}{x^3}+ \frac{x^3}{3}-6 \sqrt x)'=(\frac{3'x^3-3(x^3)'}{(x^3)^2})+ (\frac{(x^3)'3-x^3 \cdot 3'}{3^2})-$
$\displaystyle - (6' \sqrt x+6( \sqrt x)')= (\frac{0-9x^2}{x^6})+( \frac{9x^2+0}{9})- (0+ \frac{6}{2 \sqrt x})=$
$\displaystyle = - \frac{9}{x^4}+x^2- \frac{3 \sqrt x}{x}= \boxed{\frac{x^6-3x^{3,5}-9}{x^4}}$