Question

найти дифференциал (можно полностью решение!)

Answer 1

$y=xcos(3x^3), \ dy=(xcos(3x^3))'dx=(x'cos(3x^3)+x(cos(3x^3))')dx= \ =(cos(3x^3)-xsin(3x^3)cdot(3x^3)')dx= \ =(cos(3x^3)-xsin(3x^3)cdot(9x^2))dx=(cos(3x^3)-9x^3sin(3x^3))dx;$
$y= frac{sin^4x}{sqrt{2x}}, \ dy = (frac{sin^4x}{sqrt{2x}})'dx = frac{(sin^4x)'sqrt{2x}-sin^4x(sqrt{2x})'}{(sqrt{2x})^2}dx = \ = frac{4sin^3x(sin x)'sqrt{2x}-sin^4xcdot frac{sqrt{2}}{2sqrt{x}}}{2x}dx = frac{4sin^3xcos xsqrt{2x}-sin^4xcdot frac{1}{sqrt{2x}}}{2x}dx.$