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本题是隐函数的求导,具体步骤如下:
ln(x^2+y)=x^3y+sinx
(2x+y')/(x^2+y)=3x^2y+x^3y'+cosx
2x+y'=3x^2y(x^2+y)+x^3y'(x^2+y)+cosx(x^2+y)
2x+y'=3x^2y(x^2+y)+(x^5+x^3y)y'+cosx*(x^2+y)
y'[1-(x^5+x^3y)]=cosx*(x^2+y)+3x^2y(x^2+y)-2x
所以:y'=[cosx*(x^2+y)+3x^2y(x^2+y)-2x]/[1-(x^5+x^3y)],
继续对x求导,有:
y‘’
={[-sinx*(x^2+y)+cosx(2x+y')+12x^3y+3x^4y'+6xy^2+6x^2*y8y'-2][1-(x^5+x^3y)]+[cosx*(x^2+y)+3x^2y(x^2+y)-2x](5x^4+3x^2y+x^3y')}/[1-(x^5+x^3y)]^2。
ln(x^2+y)=x^3y+sinx
(2x+y')/(x^2+y)=3x^2y+x^3y'+cosx
2x+y'=3x^2y(x^2+y)+x^3y'(x^2+y)+cosx(x^2+y)
2x+y'=3x^2y(x^2+y)+(x^5+x^3y)y'+cosx*(x^2+y)
y'[1-(x^5+x^3y)]=cosx*(x^2+y)+3x^2y(x^2+y)-2x
所以:y'=[cosx*(x^2+y)+3x^2y(x^2+y)-2x]/[1-(x^5+x^3y)],
继续对x求导,有:
y‘’
={[-sinx*(x^2+y)+cosx(2x+y')+12x^3y+3x^4y'+6xy^2+6x^2*y8y'-2][1-(x^5+x^3y)]+[cosx*(x^2+y)+3x^2y(x^2+y)-2x](5x^4+3x^2y+x^3y')}/[1-(x^5+x^3y)]^2。
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