関数 $y = (x^3 + 2x)(x^2 + x + 2)(3x + 1)$ を微分して $y'$ を求める。解析学微分積の微分多項式2025/7/311. 問題の内容関数 y=(x3+2x)(x2+x+2)(3x+1)y = (x^3 + 2x)(x^2 + x + 2)(3x + 1)y=(x3+2x)(x2+x+2)(3x+1) を微分して y′y'y′ を求める。2. 解き方の手順3つの関数の積の微分を計算します。u=x3+2xu = x^3 + 2xu=x3+2x, v=x2+x+2v = x^2 + x + 2v=x2+x+2, w=3x+1w = 3x + 1w=3x+1 とおくと、y=uvwy = uvwy=uvw です。積の微分公式は、(uvw)′=u′vw+uv′w+uvw′(uvw)' = u'vw + uv'w + uvw'(uvw)′=u′vw+uv′w+uvw′です。まず、それぞれの関数を微分します。u′=3x2+2u' = 3x^2 + 2u′=3x2+2v′=2x+1v' = 2x + 1v′=2x+1w′=3w' = 3w′=3これらを積の微分公式に代入します。y′=(3x2+2)(x2+x+2)(3x+1)+(x3+2x)(2x+1)(3x+1)+(x3+2x)(x2+x+2)(3)y' = (3x^2 + 2)(x^2 + x + 2)(3x + 1) + (x^3 + 2x)(2x + 1)(3x + 1) + (x^3 + 2x)(x^2 + x + 2)(3)y′=(3x2+2)(x2+x+2)(3x+1)+(x3+2x)(2x+1)(3x+1)+(x3+2x)(x2+x+2)(3)あとはこれを展開して整理します。y′=(3x2+2)(3x3+4x2+7x+2)+(x3+2x)(6x2+5x+1)+(x3+2x)(3x2+3x+6)y' = (3x^2 + 2)(3x^3 + 4x^2 + 7x + 2) + (x^3 + 2x)(6x^2 + 5x + 1) + (x^3 + 2x)(3x^2 + 3x + 6)y′=(3x2+2)(3x3+4x2+7x+2)+(x3+2x)(6x2+5x+1)+(x3+2x)(3x2+3x+6)y′=(9x5+12x4+21x3+6x2+6x3+8x2+14x+4)+(6x5+5x4+x3+12x3+10x2+2x)+(3x5+3x4+6x3+6x3+6x2+12x)y' = (9x^5 + 12x^4 + 21x^3 + 6x^2 + 6x^3 + 8x^2 + 14x + 4) + (6x^5 + 5x^4 + x^3 + 12x^3 + 10x^2 + 2x) + (3x^5 + 3x^4 + 6x^3 + 6x^3 + 6x^2 + 12x)y′=(9x5+12x4+21x3+6x2+6x3+8x2+14x+4)+(6x5+5x4+x3+12x3+10x2+2x)+(3x5+3x4+6x3+6x3+6x2+12x)y′=9x5+12x4+27x3+14x2+14x+4+6x5+5x4+13x3+10x2+2x+3x5+3x4+12x3+6x2+12xy' = 9x^5 + 12x^4 + 27x^3 + 14x^2 + 14x + 4 + 6x^5 + 5x^4 + 13x^3 + 10x^2 + 2x + 3x^5 + 3x^4 + 12x^3 + 6x^2 + 12xy′=9x5+12x4+27x3+14x2+14x+4+6x5+5x4+13x3+10x2+2x+3x5+3x4+12x3+6x2+12xy′=(9+6+3)x5+(12+5+3)x4+(27+13+12)x3+(14+10+6)x2+(14+2+12)x+4y' = (9 + 6 + 3)x^5 + (12 + 5 + 3)x^4 + (27 + 13 + 12)x^3 + (14 + 10 + 6)x^2 + (14 + 2 + 12)x + 4y′=(9+6+3)x5+(12+5+3)x4+(27+13+12)x3+(14+10+6)x2+(14+2+12)x+4y′=18x5+20x4+52x3+30x2+28x+4y' = 18x^5 + 20x^4 + 52x^3 + 30x^2 + 28x + 4y′=18x5+20x4+52x3+30x2+28x+43. 最終的な答えy′=18x5+20x4+52x3+30x2+28x+4y' = 18x^5 + 20x^4 + 52x^3 + 30x^2 + 28x + 4y′=18x5+20x4+52x3+30x2+28x+4