$y = \sin x \cos 2x$ を微分せよ。解析学微分三角関数積の微分合成関数の微分加法定理2025/7/261. 問題の内容y=sinxcos2xy = \sin x \cos 2xy=sinxcos2x を微分せよ。2. 解き方の手順積の微分公式 (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′ を用いる。ここで u=sinxu = \sin xu=sinx, v=cos2xv = \cos 2xv=cos2x とおく。u′=(sinx)′=cosxu' = (\sin x)' = \cos xu′=(sinx)′=cosxv′=(cos2x)′=−2sin2xv' = (\cos 2x)' = -2\sin 2xv′=(cos2x)′=−2sin2xしたがって、y′=(sinxcos2x)′=(sinx)′cos2x+sinx(cos2x)′y' = (\sin x \cos 2x)' = (\sin x)' \cos 2x + \sin x (\cos 2x)'y′=(sinxcos2x)′=(sinx)′cos2x+sinx(cos2x)′=cosxcos2x+sinx(−2sin2x)= \cos x \cos 2x + \sin x (-2\sin 2x)=cosxcos2x+sinx(−2sin2x)=cosxcos2x−2sinxsin2x= \cos x \cos 2x - 2\sin x \sin 2x=cosxcos2x−2sinxsin2xここで、sin2x=2sinxcosx\sin 2x = 2\sin x \cos xsin2x=2sinxcosx, cos2x=cos2x−sin2x\cos 2x = \cos^2 x - \sin^2 xcos2x=cos2x−sin2x を用いると、y′=cosx(cos2x−sin2x)−2sinx(2sinxcosx)y' = \cos x (\cos^2 x - \sin^2 x) - 2\sin x (2\sin x \cos x)y′=cosx(cos2x−sin2x)−2sinx(2sinxcosx)=cos3x−sin2xcosx−4sin2xcosx= \cos^3 x - \sin^2 x \cos x - 4\sin^2 x \cos x=cos3x−sin2xcosx−4sin2xcosx=cos3x−5sin2xcosx= \cos^3 x - 5\sin^2 x \cos x=cos3x−5sin2xcosx=cos3x−5(1−cos2x)cosx= \cos^3 x - 5(1-\cos^2 x)\cos x=cos3x−5(1−cos2x)cosx=cos3x−5cosx+5cos3x= \cos^3 x - 5\cos x + 5\cos^3 x=cos3x−5cosx+5cos3x=6cos3x−5cosx= 6\cos^3 x - 5\cos x=6cos3x−5cosx=cosx(6cos2x−5)= \cos x(6\cos^2 x - 5)=cosx(6cos2x−5)また、y′=cosxcos2x−2sinxsin2xy' = \cos x \cos 2x - 2\sin x \sin 2xy′=cosxcos2x−2sinxsin2x に対して、和積の公式を用いることもできる。cosαcosβ=12[cos(α+β)+cos(α−β)]\cos \alpha \cos \beta = \frac{1}{2} [\cos(\alpha+\beta) + \cos(\alpha-\beta)]cosαcosβ=21[cos(α+β)+cos(α−β)]sinαsinβ=12[cos(α−β)−cos(α+β)]\sin \alpha \sin \beta = \frac{1}{2} [\cos(\alpha-\beta) - \cos(\alpha+\beta)]sinαsinβ=21[cos(α−β)−cos(α+β)]を用いると、cosxcos2x=12(cos3x+cos(−x))=12(cos3x+cosx)\cos x \cos 2x = \frac{1}{2} (\cos 3x + \cos(-x)) = \frac{1}{2} (\cos 3x + \cos x)cosxcos2x=21(cos3x+cos(−x))=21(cos3x+cosx)sinxsin2x=12(cosx−cos3x)\sin x \sin 2x = \frac{1}{2} (\cos x - \cos 3x)sinxsin2x=21(cosx−cos3x)したがって、y′=12(cos3x+cosx)−212(cosx−cos3x)y' = \frac{1}{2} (\cos 3x + \cos x) - 2\frac{1}{2} (\cos x - \cos 3x)y′=21(cos3x+cosx)−221(cosx−cos3x)=12cos3x+12cosx−cosx+cos3x= \frac{1}{2} \cos 3x + \frac{1}{2} \cos x - \cos x + \cos 3x=21cos3x+21cosx−cosx+cos3x=32cos3x−12cosx= \frac{3}{2} \cos 3x - \frac{1}{2} \cos x=23cos3x−21cosx=12(3cos3x−cosx)= \frac{1}{2} (3\cos 3x - \cos x)=21(3cos3x−cosx)=12[3(4cos3x−3cosx)−cosx]= \frac{1}{2} [3(4\cos^3 x - 3\cos x) - \cos x]=21[3(4cos3x−3cosx)−cosx]=12(12cos3x−9cosx−cosx)= \frac{1}{2} (12\cos^3 x - 9\cos x - \cos x)=21(12cos3x−9cosx−cosx)=12(12cos3x−10cosx)= \frac{1}{2} (12\cos^3 x - 10\cos x)=21(12cos3x−10cosx)=6cos3x−5cosx= 6\cos^3 x - 5\cos x=6cos3x−5cosx3. 最終的な答えy′=6cos3x−5cosxy' = 6\cos^3 x - 5\cos xy′=6cos3x−5cosx