Question

x-radical din 2x derivat

Answer 1

$(x \sqrt{2x} )'= (x)' \sqrt{2x} +x( \sqrt{2x} )'= \sqrt{2x} +x \frac{2}{2 \sqrt{2x} } =\sqrt{2x}+ \frac{2x \sqrt{2x} }{4x} =$
$= \frac{2 \sqrt{2x}+ \sqrt{2x} }{2} = \frac{3}{2} \sqrt{2x}$

Answer 2

$(x- \sqrt{2x})'=1- \frac{2x'}{2 \sqrt{2x} }=1- \frac{1}{2x}$