Question

请问这题隐函数求导的题目怎么做？

请问这题隐函数求导的题目怎么做？需要具体的过程，谢谢！

Answer 1

• 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)