$y = (x \log x - x)^2$ を微分せよ。解析学微分合成関数の微分積の微分法対数関数2025/7/301. 問題の内容y=(xlogx−x)2y = (x \log x - x)^2y=(xlogx−x)2 を微分せよ。2. 解き方の手順まず、yyyをxxxで微分します。合成関数の微分を用いると、dydx=2(xlogx−x)⋅ddx(xlogx−x)\frac{dy}{dx} = 2(x \log x - x) \cdot \frac{d}{dx}(x \log x - x)dxdy=2(xlogx−x)⋅dxd(xlogx−x)となります。次に、ddx(xlogx−x)\frac{d}{dx}(x \log x - x)dxd(xlogx−x)を計算します。積の微分法を用いると、ddx(xlogx)=ddx(x)⋅logx+x⋅ddx(logx)=1⋅logx+x⋅1x=logx+1\frac{d}{dx}(x \log x) = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1dxd(xlogx)=dxd(x)⋅logx+x⋅dxd(logx)=1⋅logx+x⋅x1=logx+1したがって、ddx(xlogx−x)=ddx(xlogx)−ddx(x)=logx+1−1=logx\frac{d}{dx}(x \log x - x) = \frac{d}{dx}(x \log x) - \frac{d}{dx}(x) = \log x + 1 - 1 = \log xdxd(xlogx−x)=dxd(xlogx)−dxd(x)=logx+1−1=logxこれらを組み合わせると、dydx=2(xlogx−x)⋅logx=2x(logx−1)logx=2xlogx(logx−1)\frac{dy}{dx} = 2(x \log x - x) \cdot \log x = 2x(\log x - 1) \log x = 2x \log x (\log x - 1)dxdy=2(xlogx−x)⋅logx=2x(logx−1)logx=2xlogx(logx−1)3. 最終的な答え2xlogx(logx−1)2x \log x (\log x - 1)2xlogx(logx−1)