设函数y=y(x)由方程sin(xy)3x-y=1所确定,求y‘(0)
sin(xy) * 3x - y = 1,x = 0,y = -1;
两边求导,3xcos(xy)( y + xy' ) + 3sin(xy) - y' = 0
[ 3x^2 * cos(xy) - 1 ]y' = -3[ xycos(xy) + sin(xy) ]
y' = -3[ xycos(xy) + sin(xy) ]/[ 3x^2 * cos(xy) - 1 ]
y'(0) = -3[ 0 + 0 ]/[ 0 - 1 ] = 0 。
据题设→y′㏑x+y/x-[(cosxy)(y+xy′)]→
y′=y[cosxy-1/x]/[㏑x-xcosxy]=(-y/x)[(1-xcosxy)/(㏑x-xcosxy)]。
sin(xy)+3x-y=1,x=0时,sin0+0-y=1,y=1。
(xdy+ydx)cos(xy)+3dx-dy=0,
[xcos(xy)-1]dy=-[ycos(xy)+3]dx,
y'=dy/dx=-[yccos(xy)+3]/[xcos(xy)-1]。
y'(0)=-(1cos0+3)/(0xos0-1)=4。
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