 | > それぞれについてdy/dxを求めよ。
> y=√{(1+sinx)/(1-sinx)}
=√{(1+sinx)^2/(1-sinx)(1+sinx)}
=√{(1+sinx)^2/(cos^2x)}=(1+sinx)/|cosx|
(cosx≧0のとき)
y'={cos^2x-(1+sinx)(-sinx)}/cos^2x={1+sinx}/{1-sin^2x}=1/(1-sinx)
(cosx<0のとき)
y'=-1/(1-sinx)
>x=e^(-√3θ)cos{2θ+(π/3)} y=e^(-√3θ)sin{2θ+(π/3)}
dx/dθ=(-√3)e^(-√3θ)cos{2θ+(π/3)}-2e^(-√3θ)sin{2θ+(π/3)}
=-√7e^(-√3θ)sin{2θ+(π/3+α)}
dy/dθ=(-√3)e^(-√3θ)sin{2θ+(π/3)}+2e^(-√3θ)cos{2θ+(π/3)}
=√7e^(-√3θ)cos{2θ+(π/3+α)}}
dy/dx=(dy/dθ)/(dx/dθ)=cos{2θ+(π/3+α)}/sin{2θ+(π/3+α)}
=-tan(2θ+(α-π/6)) (∵tanα=√3/2)
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