求二元函数的偏倒数.1.z=xy+ln(x^2+y^2) 2.z=cos^2(x+2y)(二阶)

1个回答

  • 1,z'(x)=y+2x/(x^2+y^2);z'(y)=x+2y/(x^2+y^2).

    2.z'(x)=2cos(x+2y)(-sin(x+2y)=-sin(2x+4y);z'(y)=4cos(x+2y)(-sin(x+2y)=-2sin(2x+4y);

    z''(xx)=-2cos(2x+4y);z''(xy)=z''(yx)=-4cos(2x+4y);z''(yy)=-8cos(2x+4y)..

    三.应用题

    1.联立x=y^2 y=x-2得交点为(1,-1),(4,2).积分区域为:-1《y《2,y^2《x《y+2

    所以:面积S=∫ (-1,2)dy∫ (y^2,y+2)dx=∫ (-1,2) ( y+2 - y^2) dy=[y^2/2+2y-y^3/3] |(-1,2)=9/2

    2.这是一阶线性微分方程.y‘-2xy=3x的通解为:y=e^(2x)(∫ 3xe^(-2x)dx+C)

    =Ce^(2x)-3/2