高数 极坐标弧长积分 请问 ds=根号（dx^2+dy^2）是怎么推出根号（r(θ)^2+r’（θ）

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ds=√[(dx)²+(dy)²]=√[(dx)²+(y')²(dx)²]=√[1+(y')²]dx
x=r(θ)cosθ,y=r(θ)sinθ
√[1+(y')²]dx=√[1+(d(rsinθ)/dx)²]dx
=√[1+((d(rsinθ)/dθ)*dθ/dx)²]*(dx/dθ)dθ
=√[(dx/dθ)²+(d(r(θ)sinθ)/dθ)²]dθ
=√[(dx/dθ)²+(r'(θ)sinθ+r(θ)cosθ)²]dθ
=√[(r'(θ)cosθ-r(θ)sinθ)²+(r'(θ)sinθ+r(θ)cosθ)²]dθ
=√[(r'(θ))²+(r(θ))²]dθ
======以下答案可供参考======
供参考答案1:
直角坐标与极坐标的关系x＝r(θ)cosθ，y=r(θ)sinθ
dx/dθ＝r'(θ)cosθ－r(θ)sinθ
dy/dθ＝r'(θ)sinθ＋r(θ)cosθ
(dx/dθ)^2＋(dy/dθ)^2＝[r'(θ)]^2＋[r(θ)]^2
ds=√[(dx)²+(dy)²]=√[(dx/dθ)²+(dy/dθ)²]dθ＝√((r'(θ))^2＋(r(θ))^2)dθ