已知ysinx-cos(x+y)=0,求在点(0,π/2)的dy/dx值
人气:358 ℃ 时间:2020-01-25 04:49:33
解答
ysinx-cos(x+y)=0,两边对x求导,得
y'sinx+ycosx+(1+y')sin(x+y)=0,解得
y'=-[ycosx+sin(x+y)]/[sinx+sin(x+y)]
dy/dx=y'(0,π/2)=-(π/2+1)/(0+1)=-(π/2+1)
推荐
- 已知ysinx-cos(x+y)=0,求在点(0,π)的dy/dx值
- ysinx+cos(x-y)=0,求dy/dx|(x=π/2)
- ysinx-cos(x+y)=0,求 dy/dx
- 求由方程ysinx-cos(xy)=0所确定的隐函数y=y(x)的导数dy/dx
- 求通解:y'=x(1+y^2);sinxdy=2ycosxdx; y^2 dx -(xy+1)dy=0;求特解:cosx(dy/dx)+ysinx=cos^2x,x=∏时y=1
- 氨气溶于水的化学方程式
- 银行定期存款到期后,不取自动转存,自动转存后没到期就取钱,利息怎样算
- 急!【高一数学】已知定义域为R的函数f(x)在(2,+∞)上为增函数,且函数y=f(x+2)为偶函数
猜你喜欢