y=y(x)由方程 [e^(x+y)]+sin(xy)=1确定,求y'(x)及y'(0)

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y=y(x)由方程 [e^(x+y)]+sin(xy)=1确定,求y'(x)及y'(0)
y=y(x)由方程 [e^(x+y)]+sin(xy)=1确定,求y'(x)及y'(0)

y=y(x)由方程 [e^(x+y)]+sin(xy)=1确定,求y'(x)及y'(0)
e^(x+y) + sin(xy) = 1
e^(x+y)*(1+y')+cos(xy)(y+xy')=0
y'*[e*(x+y)+xcos(xy)]=-[ycos(xy)+e^(x+y)]
y'=-[ycos(xy)+e^(x+y)]/[e*(x+y)+xcos(xy)]
x=0,求出 y=0,
代入上式,得到y'（x=0）=-1.

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