関数 $y = (x^2 - 1)(1 - x^4)$ を微分せよ。解析学微分関数の微分多項式2025/7/51. 問題の内容関数 y=(x2−1)(1−x4)y = (x^2 - 1)(1 - x^4)y=(x2−1)(1−x4) を微分せよ。2. 解き方の手順まず、yyy を展開します。y=(x2−1)(1−x4)=x2−x6−1+x4=−x6+x4+x2−1y = (x^2 - 1)(1 - x^4) = x^2 - x^6 - 1 + x^4 = -x^6 + x^4 + x^2 - 1y=(x2−1)(1−x4)=x2−x6−1+x4=−x6+x4+x2−1次に、各項を微分します。dydx=ddx(−x6+x4+x2−1)\frac{dy}{dx} = \frac{d}{dx}(-x^6 + x^4 + x^2 - 1)dxdy=dxd(−x6+x4+x2−1)ddx(−x6)=−6x5\frac{d}{dx}(-x^6) = -6x^5dxd(−x6)=−6x5ddx(x4)=4x3\frac{d}{dx}(x^4) = 4x^3dxd(x4)=4x3ddx(x2)=2x\frac{d}{dx}(x^2) = 2xdxd(x2)=2xddx(−1)=0\frac{d}{dx}(-1) = 0dxd(−1)=0したがって、dydx=−6x5+4x3+2x\frac{dy}{dx} = -6x^5 + 4x^3 + 2xdxdy=−6x5+4x3+2x3. 最終的な答えdydx=−6x5+4x3+2x\frac{dy}{dx} = -6x^5 + 4x^3 + 2xdxdy=−6x5+4x3+2x