与えられた関数 $y = (x^3 + 1)(x + 1)(x^2 + x - 2)$ を微分して、$dy/dx$を求める問題です。解析学微分関数の微分多項式2025/7/51. 問題の内容与えられた関数 y=(x3+1)(x+1)(x2+x−2)y = (x^3 + 1)(x + 1)(x^2 + x - 2)y=(x3+1)(x+1)(x2+x−2) を微分して、dy/dxdy/dxdy/dxを求める問題です。2. 解き方の手順まず、x2+x−2x^2 + x - 2x2+x−2を因数分解します。x2+x−2=(x+2)(x−1)x^2 + x - 2 = (x + 2)(x - 1)x2+x−2=(x+2)(x−1)次に、x3+1x^3 + 1x3+1を因数分解します。x3+1=(x+1)(x2−x+1)x^3 + 1 = (x + 1)(x^2 - x + 1)x3+1=(x+1)(x2−x+1)したがって、yyyは次のように書き換えられます。y=(x+1)(x2−x+1)(x+1)(x+2)(x−1)y = (x + 1)(x^2 - x + 1)(x + 1)(x + 2)(x - 1)y=(x+1)(x2−x+1)(x+1)(x+2)(x−1)y=(x+1)2(x2−x+1)(x+2)(x−1)y = (x + 1)^2 (x^2 - x + 1) (x + 2)(x - 1)y=(x+1)2(x2−x+1)(x+2)(x−1)さらに、yyyを展開します。y=(x2+2x+1)(x2−x+1)(x2+x−2)y = (x^2 + 2x + 1)(x^2 - x + 1)(x^2 + x - 2)y=(x2+2x+1)(x2−x+1)(x2+x−2)y=(x4+x3−x2+x+1)(x+2)(x−1)y = (x^4 + x^3 - x^2 + x + 1)(x + 2)(x-1)y=(x4+x3−x2+x+1)(x+2)(x−1)y=(x4+x3+1)(x+1)(x2+x−2)=(x4+x3+1)(x3+2x2−x2−2x+x+2)=(x4+x3+1)(x3+x2−x−2)y = (x^4 + x^3 + 1)(x+1)(x^2+x-2)=(x^4 + x^3 + 1)(x^3 + 2x^2 - x^2 -2x + x + 2)= (x^4 + x^3 + 1)(x^3 + x^2 - x - 2)y=(x4+x3+1)(x+1)(x2+x−2)=(x4+x3+1)(x3+2x2−x2−2x+x+2)=(x4+x3+1)(x3+x2−x−2)y=x7+x6−x5−2x4+x6+x5−x4−2x3+x3+x2−x−2y = x^7 + x^6 - x^5 - 2x^4 + x^6 + x^5 - x^4 - 2x^3 + x^3 + x^2 - x - 2y=x7+x6−x5−2x4+x6+x5−x4−2x3+x3+x2−x−2y=x7+2x6−3x4−x3+x2−x−2y = x^7 + 2x^6 - 3x^4 - x^3 + x^2 - x - 2y=x7+2x6−3x4−x3+x2−x−2ここで、べき乗のルールを使用して微分します。ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}dxd(xn)=nxn−1dydx=7x6+12x5−12x3−3x2+2x−1\frac{dy}{dx} = 7x^6 + 12x^5 - 12x^3 - 3x^2 + 2x - 1dxdy=7x6+12x5−12x3−3x2+2x−13. 最終的な答えdydx=7x6+12x5−12x3−3x2+2x−1\frac{dy}{dx} = 7x^6 + 12x^5 - 12x^3 - 3x^2 + 2x - 1dxdy=7x6+12x5−12x3−3x2+2x−1