求二阶导（y',y''每一步每个数据都请详细,1.y=ln(sinx) 2.y=(e^x^2)*cosx 3.y=sin

y=ln(sinx) y'=(1/sinx)dsinx=cosx/sinx y"=-1/sin^2x
y=(e^x^2)*cosx y'=(e^x^2)*2x*cosx-(e^x^2)*sinx y"=2(e^x^2)*cosx+4x^2*(e^x^2)*cosx-(e^x^2)*2x*sinx-[(e^x^2)*2x*sinx+(e^x^2)*cosx]
y=sin(x+y) y'=cos(x+y)(1+y') => y'=cos(x+y)/(1-cos(x+y)) y"=-sin(x+y)[1+cos(x+y)]/[1-cos(x+y)]^3(步骤很麻烦就不详写了,这些基本的还是得多练练)