Question

Найти y'(x)
$y= frac{x^2}{x+1} \ f(x)=(x^2+3x)* sqrt{x}$

Answer 1

$y= frac{x^2}{x+1} \ y'= frac{2x(x+1)-x^2}{(x+1)^2}= frac{x^2+2x}{(x+1)^2}$

$f(x)=(x^2+3x) sqrt{x} =(x^2 sqrt{x} +3x sqrt{x} ) \ f'(x)= frac{5x sqrt{x} +9 sqrt{x} }{2}$