Найти производные 2 порядка: у=1/х * lgx y=sin2x+cos2x

[latex]y=\frac{1}{x}\cdot lgx\\\\y'=-\frac{1}{x^2}\cdot lgx+\frac{1}{x}\cdot \frac{1}{x}=\frac{1}{x^2}(1-lgx)\\\\y''=-\frac{2}{x^3}\cdot (1-lgx)+\frac{1}{x^2}(-\frac{1}{x})=-\frac{1}{x^3}(2-2lgx+1)=-\frac{1}{x^3}(3-2lgx)[/latex] [latex]2)\; \; y=sin2x+cos2x\\\\y'=2cos2x-2sin2x=2(cos2x-sin2x)\\\\y''=2(-2sin2x-2cos2x)=-4(sin2x+cos2x)[/latex]