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y=f(x)由方程sin(xy)-ln(x+y)/y=1确定,求y的导数
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y=f(x)由方程sin(xy)-ln(x+y)/y=1确定,求y的导数
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答案和解析
y sin(xy) - ln(x+y) = y
y' sin(xy) + y * cos(xy) * (y + x*y') - (1+y') / (x+y) = y'
y' sin(xy) + xycos(xy) y' - y'/(x+y) - y' = 1/(x+y) - y^2 cos(xy)
y' = [1/(x+y) - y^2 cos(xy)] / [sin(xy) + xycos(xy) - 1/(x+y) - 1]
dy/dx = [1 - (x+y) y^2 cos(xy)] / [(x+y)sin(xy) + xy(x+y)cos(xy) - x - y - 1]
y' sin(xy) + y * cos(xy) * (y + x*y') - (1+y') / (x+y) = y'
y' sin(xy) + xycos(xy) y' - y'/(x+y) - y' = 1/(x+y) - y^2 cos(xy)
y' = [1/(x+y) - y^2 cos(xy)] / [sin(xy) + xycos(xy) - 1/(x+y) - 1]
dy/dx = [1 - (x+y) y^2 cos(xy)] / [(x+y)sin(xy) + xy(x+y)cos(xy) - x - y - 1]
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