次の関数を微分する問題です。 (1) $y = \sin^{-1} \frac{x}{\sqrt{1+x^2}}$ (2) $y = \tan^{-1} \frac{1-\cos x}{\sin x}$解析学微分逆三角関数置換積分2025/7/91. 問題の内容次の関数を微分する問題です。(1) y=sin−1x1+x2y = \sin^{-1} \frac{x}{\sqrt{1+x^2}}y=sin−11+x2x(2) y=tan−11−cosxsinxy = \tan^{-1} \frac{1-\cos x}{\sin x}y=tan−1sinx1−cosx2. 解き方の手順(1) y=sin−1x1+x2y = \sin^{-1} \frac{x}{\sqrt{1+x^2}}y=sin−11+x2xx=tanθx = \tan \thetax=tanθ と置換します。このとき、dx=sec2θdθdx = \sec^2 \theta d\thetadx=sec2θdθです。x1+x2=tanθ1+tan2θ=tanθsec2θ=tanθsecθ=sinθ/cosθ1/cosθ=sinθ\frac{x}{\sqrt{1+x^2}} = \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} = \frac{\tan \theta}{\sqrt{\sec^2 \theta}} = \frac{\tan \theta}{\sec \theta} = \frac{\sin \theta / \cos \theta}{1 / \cos \theta} = \sin \theta1+x2x=1+tan2θtanθ=sec2θtanθ=secθtanθ=1/cosθsinθ/cosθ=sinθしたがって、y=sin−1(sinθ)=θy = \sin^{-1}(\sin \theta) = \thetay=sin−1(sinθ)=θ。x=tanθx = \tan \thetax=tanθ より、θ=tan−1x\theta = \tan^{-1} xθ=tan−1x。よって、y=tan−1xy = \tan^{-1} xy=tan−1x。dydx=ddx(tan−1x)=11+x2\frac{dy}{dx} = \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}dxdy=dxd(tan−1x)=1+x21(2) y=tan−11−cosxsinxy = \tan^{-1} \frac{1-\cos x}{\sin x}y=tan−1sinx1−cosx1−cosxsinx=2sin2(x/2)2sin(x/2)cos(x/2)=sin(x/2)cos(x/2)=tan(x/2)\frac{1-\cos x}{\sin x} = \frac{2 \sin^2 (x/2)}{2 \sin (x/2) \cos (x/2)} = \frac{\sin (x/2)}{\cos (x/2)} = \tan (x/2)sinx1−cosx=2sin(x/2)cos(x/2)2sin2(x/2)=cos(x/2)sin(x/2)=tan(x/2)したがって、y=tan−1(tan(x/2))=x/2y = \tan^{-1}(\tan (x/2)) = x/2y=tan−1(tan(x/2))=x/2。dydx=ddx(x/2)=12\frac{dy}{dx} = \frac{d}{dx} (x/2) = \frac{1}{2}dxdy=dxd(x/2)=213. 最終的な答え(1) dydx=11+x2\frac{dy}{dx} = \frac{1}{1+x^2}dxdy=1+x21(2) dydx=12\frac{dy}{dx} = \frac{1}{2}dxdy=21