z=(2x+y)sin(x+y^2)对x求偏导
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z=(2x+y)sin(x+y^2)
∂z/∂x
=∂/∂x[(2x+y)sin(x+y^2)]
=(2x+y) ∂/∂x[ sin(x+y^2)] +sin(x+y^2).∂/∂x(2x+y)
=(2x+y).cos(x+y^2).∂/∂x(x+y^2) +sin(x+y^2).(2)
=(2x+y).cos(x+y^2).(1) +2sin(x+y^2)
=(2x+y).cos(x+y^2) +2sin(x+y^2)
∂z/∂x
=∂/∂x[(2x+y)sin(x+y^2)]
=(2x+y) ∂/∂x[ sin(x+y^2)] +sin(x+y^2).∂/∂x(2x+y)
=(2x+y).cos(x+y^2).∂/∂x(x+y^2) +sin(x+y^2).(2)
=(2x+y).cos(x+y^2).(1) +2sin(x+y^2)
=(2x+y).cos(x+y^2) +2sin(x+y^2)
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