関数 $y = (x^3 - 3x^2 + 5)(2x + 1)$ を微分せよ。解析学微分積の微分多項式2025/7/231. 問題の内容関数 y=(x3−3x2+5)(2x+1)y = (x^3 - 3x^2 + 5)(2x + 1)y=(x3−3x2+5)(2x+1) 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′ を使用します。まず、u=x3−3x2+5u = x^3 - 3x^2 + 5u=x3−3x2+5 と v=2x+1v = 2x + 1v=2x+1 とおきます。次に、uuu と vvv をそれぞれ微分します。u′=ddx(x3−3x2+5)=3x2−6xu' = \frac{d}{dx}(x^3 - 3x^2 + 5) = 3x^2 - 6xu′=dxd(x3−3x2+5)=3x2−6xv′=ddx(2x+1)=2v' = \frac{d}{dx}(2x + 1) = 2v′=dxd(2x+1)=2積の微分公式に代入します。y′=u′v+uv′=(3x2−6x)(2x+1)+(x3−3x2+5)(2)y' = u'v + uv' = (3x^2 - 6x)(2x + 1) + (x^3 - 3x^2 + 5)(2)y′=u′v+uv′=(3x2−6x)(2x+1)+(x3−3x2+5)(2)これを展開して整理します。y′=(6x3+3x2−12x2−6x)+(2x3−6x2+10)y' = (6x^3 + 3x^2 - 12x^2 - 6x) + (2x^3 - 6x^2 + 10)y′=(6x3+3x2−12x2−6x)+(2x3−6x2+10)y′=6x3+3x2−12x2−6x+2x3−6x2+10y' = 6x^3 + 3x^2 - 12x^2 - 6x + 2x^3 - 6x^2 + 10y′=6x3+3x2−12x2−6x+2x3−6x2+10y′=(6x3+2x3)+(3x2−12x2−6x2)−6x+10y' = (6x^3 + 2x^3) + (3x^2 - 12x^2 - 6x^2) - 6x + 10y′=(6x3+2x3)+(3x2−12x2−6x2)−6x+10y′=8x3−15x2−6x+10y' = 8x^3 - 15x^2 - 6x + 10y′=8x3−15x2−6x+103. 最終的な答えy′=8x3−15x2−6x+10y' = 8x^3 - 15x^2 - 6x + 10y′=8x3−15x2−6x+10