Question

Y=корень2x дифференциал

Answer 1

$y = \sqrt{2x} \ ;$

$dy = d(2x)^{1/2} = \frac{1}{2} (2x)^{1/2-1} dx = \frac{1}{2} (2x)^{-1/2} dx = \frac{dx}{ 2 (2x)^{1/2} } = \frac{dx}{ 2 \sqrt{2x} } \ .$