Question

Дана функция
$f(x) = 2 sqrt{x} + frac{3}{ {x}^{2} }$
Найдите f'(x), f'(1)

Answer 1

$f(x)=2sqrt{x}+frac{3}{x^2}=2sqrt{x}+3x^{-2}\\ f'(x)=[2sqrt{x}+3x^{-2}]'=[2sqrt{x}]'+[3x^{-2}]'=\\ =2*[sqrt{x}]'+3*[x^{-2}]'=2*frac{1}{2sqrt{x}}+3*(-2)*x^{-3}=\\ =frac{1}{sqrt{x}}-frac{6}{x^{3}}.\\ f'(1)=frac{1}{sqrt{1}}-frac{6}{1^{3}}=1-6=-5$