Найти dy 1)y=(ax^2+b)^3 2)y=arctg корень2x

[latex]y=(ax^2+b)^3 ; \\ dy=d((ax^2+b)^3)=\\ 3(ax^2+b)^{3-1}d(ax^2+b)=\\ 3(ax^2+b)^2 d(ax^2)=\\ 3(ax^2+b)^2*a*2x dx=\\ 6ax(ax^2+b)^2 dx [/latex]   [latex]y=arctg \sqrt{2x};\\ dy=\frac {1}{1+(\sqrt{2x})^2}d (\sqrt{2x})=\\ \frac{1}{1+2x}\frac{1}{2\sqrt{2x}}d (2x)=\\ \frac{1}{2(1+2x)\sqrt{2x}} 2d x=\\ \frac{1}{(1+2x)\sqrt{2x}} dx[/latex]