関数 $y = (x^3 - x)(x^2 + 1)(3x^4 + x^2)$ を微分する問題です。解析学微分積の微分法導関数2025/7/51. 問題の内容関数 y=(x3−x)(x2+1)(3x4+x2)y = (x^3 - x)(x^2 + 1)(3x^4 + x^2)y=(x3−x)(x2+1)(3x4+x2) を微分する問題です。2. 解き方の手順積の微分法を使います。関数をu=x3−xu = x^3 - xu=x3−x, v=x2+1v = x^2 + 1v=x2+1, w=3x4+x2w = 3x^4 + x^2w=3x4+x2 とおくと、y=uvwy = uvwy=uvw です。積の微分法により、y′=u′vw+uv′w+uvw′y' = u'vw + uv'w + uvw'y′=u′vw+uv′w+uvw′ となります。まず、各関数の導関数を求めます。u′=(x3−x)′=3x2−1u' = (x^3 - x)' = 3x^2 - 1u′=(x3−x)′=3x2−1v′=(x2+1)′=2xv' = (x^2 + 1)' = 2xv′=(x2+1)′=2xw′=(3x4+x2)′=12x3+2xw' = (3x^4 + x^2)' = 12x^3 + 2xw′=(3x4+x2)′=12x3+2x次に、これらの導関数をy′=u′vw+uv′w+uvw′y' = u'vw + uv'w + uvw'y′=u′vw+uv′w+uvw′ に代入します。y′=(3x2−1)(x2+1)(3x4+x2)+(x3−x)(2x)(3x4+x2)+(x3−x)(x2+1)(12x3+2x)y' = (3x^2 - 1)(x^2 + 1)(3x^4 + x^2) + (x^3 - x)(2x)(3x^4 + x^2) + (x^3 - x)(x^2 + 1)(12x^3 + 2x)y′=(3x2−1)(x2+1)(3x4+x2)+(x3−x)(2x)(3x4+x2)+(x3−x)(x2+1)(12x3+2x)それぞれの項を展開して整理します。第1項: (3x2−1)(x2+1)(3x4+x2)=(3x4+3x2−x2−1)(3x4+x2)=(3x4+2x2−1)(3x4+x2)=9x8+3x6+6x6+2x4−3x4−x2=9x8+9x6−x4−x2(3x^2 - 1)(x^2 + 1)(3x^4 + x^2) = (3x^4 + 3x^2 - x^2 - 1)(3x^4 + x^2) = (3x^4 + 2x^2 - 1)(3x^4 + x^2) = 9x^8 + 3x^6 + 6x^6 + 2x^4 - 3x^4 - x^2 = 9x^8 + 9x^6 - x^4 - x^2(3x2−1)(x2+1)(3x4+x2)=(3x4+3x2−x2−1)(3x4+x2)=(3x4+2x2−1)(3x4+x2)=9x8+3x6+6x6+2x4−3x4−x2=9x8+9x6−x4−x2第2項: (x3−x)(2x)(3x4+x2)=(2x4−2x2)(3x4+x2)=6x8+2x6−6x6−2x4=6x8−4x6−2x4(x^3 - x)(2x)(3x^4 + x^2) = (2x^4 - 2x^2)(3x^4 + x^2) = 6x^8 + 2x^6 - 6x^6 - 2x^4 = 6x^8 - 4x^6 - 2x^4(x3−x)(2x)(3x4+x2)=(2x4−2x2)(3x4+x2)=6x8+2x6−6x6−2x4=6x8−4x6−2x4第3項: (x3−x)(x2+1)(12x3+2x)=(x5+x3−x3−x)(12x3+2x)=(x5−x)(12x3+2x)=12x8+2x6−12x4−2x2(x^3 - x)(x^2 + 1)(12x^3 + 2x) = (x^5 + x^3 - x^3 - x)(12x^3 + 2x) = (x^5 - x)(12x^3 + 2x) = 12x^8 + 2x^6 - 12x^4 - 2x^2(x3−x)(x2+1)(12x3+2x)=(x5+x3−x3−x)(12x3+2x)=(x5−x)(12x3+2x)=12x8+2x6−12x4−2x2したがって、y′=(9x8+9x6−x4−x2)+(6x8−4x6−2x4)+(12x8+2x6−12x4−2x2)y' = (9x^8 + 9x^6 - x^4 - x^2) + (6x^8 - 4x^6 - 2x^4) + (12x^8 + 2x^6 - 12x^4 - 2x^2)y′=(9x8+9x6−x4−x2)+(6x8−4x6−2x4)+(12x8+2x6−12x4−2x2)y′=(9+6+12)x8+(9−4+2)x6+(−1−2−12)x4+(−1−2)x2y' = (9 + 6 + 12)x^8 + (9 - 4 + 2)x^6 + (-1 - 2 - 12)x^4 + (-1 - 2)x^2y′=(9+6+12)x8+(9−4+2)x6+(−1−2−12)x4+(−1−2)x2y′=27x8+7x6−15x4−3x2y' = 27x^8 + 7x^6 - 15x^4 - 3x^2y′=27x8+7x6−15x4−3x23. 最終的な答えy′=27x8+7x6−15x4−3x2y' = 27x^8 + 7x^6 - 15x^4 - 3x^2y′=27x8+7x6−15x4−3x2