関数 $y = (x^3 - x)(x^2 + 1)(3x^4 + x^2)$ を微分せよ。解析学微分積の微分多項式2025/7/271. 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) を微分せよ。2. 解き方の手順積の微分法を用いる。3つの関数 u(x)u(x)u(x), v(x)v(x)v(x), w(x)w(x)w(x) の積の微分は、(uvw)′=u′vw+uv′w+uvw′(uvw)' = u'vw + uv'w + uvw'(uvw)′=u′vw+uv′w+uvw′で与えられる。ここで、u(x)=x3−xu(x) = x^3 - xu(x)=x3−xv(x)=x2+1v(x) = x^2 + 1v(x)=x2+1w(x)=3x4+x2w(x) = 3x^4 + x^2w(x)=3x4+x2とおくと、u′(x)=3x2−1u'(x) = 3x^2 - 1u′(x)=3x2−1v′(x)=2xv'(x) = 2xv′(x)=2xw′(x)=12x3+2xw'(x) = 12x^3 + 2xw′(x)=12x3+2xとなる。したがって、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)=(3x4+2x2−1)(3x4+x2)+(x3−x)(6x5+2x3)+(x5+x3−x3−x)(12x3+2x)= (3x^4 + 2x^2 - 1)(3x^4 + x^2) + (x^3 - x)(6x^5 + 2x^3) + (x^5 + x^3 - x^3 - x)(12x^3 + 2x)=(3x4+2x2−1)(3x4+x2)+(x3−x)(6x5+2x3)+(x5+x3−x3−x)(12x3+2x)=(9x8+3x6+6x6+2x4−3x4−x2)+(6x8+2x6−6x6−2x4)+(x5−x)(12x3+2x)= (9x^8 + 3x^6 + 6x^6 + 2x^4 - 3x^4 - x^2) + (6x^8 + 2x^6 - 6x^6 - 2x^4) + (x^5 - x)(12x^3 + 2x)=(9x8+3x6+6x6+2x4−3x4−x2)+(6x8+2x6−6x6−2x4)+(x5−x)(12x3+2x)=(9x8+9x6−x4−x2)+(6x8−4x6−2x4)+(12x8+2x6−12x4−2x2)= (9x^8 + 9x^6 - x^4 - x^2) + (6x^8 - 4x^6 - 2x^4) + (12x^8 + 2x^6 - 12x^4 - 2x^2)=(9x8+9x6−x4−x2)+(6x8−4x6−2x4)+(12x8+2x6−12x4−2x2)=9x8+9x6−x4−x2+6x8−4x6−2x4+12x8+2x6−12x4−2x2= 9x^8 + 9x^6 - x^4 - x^2 + 6x^8 - 4x^6 - 2x^4 + 12x^8 + 2x^6 - 12x^4 - 2x^2=9x8+9x6−x4−x2+6x8−4x6−2x4+12x8+2x6−12x4−2x2=(9+6+12)x8+(9−4+2)x6+(−1−2−12)x4+(−1−2)x2= (9 + 6 + 12)x^8 + (9 - 4 + 2)x^6 + (-1 - 2 - 12)x^4 + (-1 - 2)x^2=(9+6+12)x8+(9−4+2)x6+(−1−2−12)x4+(−1−2)x2=27x8+7x6−15x4−3x2= 27x^8 + 7x^6 - 15x^4 - 3x^2=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