関数 $y = x^2 e^{-x}$ の第4次導関数を求める問題です。解析学微分導関数関数の微分4次導関数2025/7/131. 問題の内容関数 y=x2e−xy = x^2 e^{-x}y=x2e−x の第4次導関数を求める問題です。2. 解き方の手順まず、与えられた関数 y=x2e−xy = x^2 e^{-x}y=x2e−x を微分していきます。積の微分法 ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv'dxd(uv)=u′v+uv′ を利用します。1階微分:y′=ddx(x2e−x)=(2x)e−x+x2(−e−x)=2xe−x−x2e−x=(2x−x2)e−xy' = \frac{d}{dx}(x^2 e^{-x}) = (2x)e^{-x} + x^2(-e^{-x}) = 2xe^{-x} - x^2e^{-x} = (2x-x^2)e^{-x}y′=dxd(x2e−x)=(2x)e−x+x2(−e−x)=2xe−x−x2e−x=(2x−x2)e−x2階微分:y′′=ddx((2x−x2)e−x)=(2−2x)e−x+(2x−x2)(−e−x)=(2−2x)e−x−(2x−x2)e−x=(2−4x+x2)e−xy'' = \frac{d}{dx}((2x-x^2)e^{-x}) = (2-2x)e^{-x} + (2x-x^2)(-e^{-x}) = (2-2x)e^{-x} - (2x-x^2)e^{-x} = (2-4x+x^2)e^{-x}y′′=dxd((2x−x2)e−x)=(2−2x)e−x+(2x−x2)(−e−x)=(2−2x)e−x−(2x−x2)e−x=(2−4x+x2)e−x3階微分:y′′′=ddx((2−4x+x2)e−x)=(−4+2x)e−x+(2−4x+x2)(−e−x)=(−4+2x)e−x−(2−4x+x2)e−x=(−6+6x−x2)e−xy''' = \frac{d}{dx}((2-4x+x^2)e^{-x}) = (-4+2x)e^{-x} + (2-4x+x^2)(-e^{-x}) = (-4+2x)e^{-x} - (2-4x+x^2)e^{-x} = (-6+6x-x^2)e^{-x}y′′′=dxd((2−4x+x2)e−x)=(−4+2x)e−x+(2−4x+x2)(−e−x)=(−4+2x)e−x−(2−4x+x2)e−x=(−6+6x−x2)e−x4階微分:y(4)=ddx((−6+6x−x2)e−x)=(6−2x)e−x+(−6+6x−x2)(−e−x)=(6−2x)e−x−(−6+6x−x2)e−x=(12−8x+x2)e−xy^{(4)} = \frac{d}{dx}((-6+6x-x^2)e^{-x}) = (6-2x)e^{-x} + (-6+6x-x^2)(-e^{-x}) = (6-2x)e^{-x} - (-6+6x-x^2)e^{-x} = (12-8x+x^2)e^{-x}y(4)=dxd((−6+6x−x2)e−x)=(6−2x)e−x+(−6+6x−x2)(−e−x)=(6−2x)e−x−(−6+6x−x2)e−x=(12−8x+x2)e−x3. 最終的な答えy(4)=(x2−8x+12)e−xy^{(4)} = (x^2 - 8x + 12)e^{-x}y(4)=(x2−8x+12)e−x