Question

Найти производную....
$y=\frac{1}{\sqrt{3x+1} }$

Answer 1

$\displaystyle \left(\frac{1}{\sqrt{3x+1}}\right)' = \left((3x+1)^{-1/2}\right)'=-\frac{1}{2}(3x+1)^{-3/2}\cdot(3x+1)' = -\frac{3}{2(3x+1)^{3/2}}$