两边对x求导
cos(xy)·(y+xy')=[y/(x+e)]·[y+(x+e)y']/y²=[y+(x+e)y']/y(x+e)
y(x+e)cos(xy)·(y+xy')=y+(x+e)y'
(xy²+ey²)cos(xy)-y=[(x+e)-(x²y+exy)cos(xy)]y'
y'=[(xy²+ey²)cos(xy)-y]/[(x+e)-(x²y+exy)cos(xy)]
sin(0·y)=ln(e/y)+1→y=e²
y'(0)=[e³-e²]/[e]=e²-e
①=(xy²+ey²)cos(xy)-y→①(0)=e³-e²
①'=(y²+2xyy'+2eyy')cos(xy)+(xy²+ey²)sin(xy)·(y+xy')→①'(0)=e⁴+2e·e²(e³-e²)
②=(x+e)-(x²y+exy)cos(xy)→②(0)=e
②'=1-(2xy+x²y'+ey+exy)cos(xy)+(x²y+exy)sin(xy)·(y+xy')→②'(0)=1-e³
y''(0)=[①'(0)·②(0)-①(0)·②'(0)]/②²(0)
=[(2e⁶-2e⁵+e⁴)·e-(e³-e²)·(1-e³)]/e²
=2e⁵-2e⁴+e³-(e-1)(1-e³)
=2e⁵+e⁴-e+1
(计算过程较繁,不知有无错漏,但计算方法应该就是这样了)
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