Calc 3 help?

First, we derivative the formula of the slope of a polar curve.
x = r cos θ = f(θ) cos θ
dx/dθ = (df/dθ)cos θ + f ( - sin θ ) = r' cos θ - r sin θ

y = r sin θ = f(θ) sin θ
dy/dθ = (df/dθ)sin θ + f cos θ = r' sin θ + r cos θ

dy/dx
= (dy/dθ) / (dx/dθ)
= ( r' sin θ + r cos θ ) / ( r' cos θ - r sin θ ) ..... Formula
where r = f(θ) , r' = df/dθ

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For this problem
r = f(θ) = 9 sin θ
r' = df/dθ = 9 cos θ

dy/dx
= ( r' sin θ + r cos θ ) / ( r' cos θ - r sin θ )
= ( 9 cos θ sin θ + 9 sin θ cos θ ) / ( 9 cos θ cos θ - 9 sin θ sin θ )
= 18 sin θ cos θ / ( 9 cos² θ - 9 sin² θ )
= 2 sin θ cos θ / ( cos² θ - sin² θ )
= sin 2θ / cos 2θ

At θ = π/6 = 30° ,

dy/dx
= sin 60° / cos 60°
= (√3 / 2 ) / (1/2)
= √3 ..... Ans