这题矩阵怎么算?求大神解答QAQ
2个回答
2017-11-07
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arctan(x/y)=0.5ln(x²+y²)
两边对x求导:
1/(1+x²/y²)* (x/y)'=0.5/(x²+y²)* (x²+y²)'
y²/(x²+y²)* (y-xy')/y²=0.5/(x²+y²)* (2x+2yy')
y-xy'=x+yy'
(y+x)y'=y-x
y'=(y-x)/(y+x)
继续对x求导:
y"=[(y'-1)(y+x)-(y-x)(y'+1)]/(y+x)²
=2(xy'-y)/(y+x)²
=2[x(y-x)/(y+x)-y]/(y+x)²
=-2(x²+y²)/(y+x)³
两边对x求导:
1/(1+x²/y²)* (x/y)'=0.5/(x²+y²)* (x²+y²)'
y²/(x²+y²)* (y-xy')/y²=0.5/(x²+y²)* (2x+2yy')
y-xy'=x+yy'
(y+x)y'=y-x
y'=(y-x)/(y+x)
继续对x求导:
y"=[(y'-1)(y+x)-(y-x)(y'+1)]/(y+x)²
=2(xy'-y)/(y+x)²
=2[x(y-x)/(y+x)-y]/(y+x)²
=-2(x²+y²)/(y+x)³
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