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今日の微分

2020/09/23に公開

三角関数の微分

\begin{aligned} \left(\frac{\sin x}{\sin x+\cos x}\right)'&=\frac{\cos x(\sin x+\cos x)-\sin x(\cos x-\sin x)}{(\sin x+\cos x)^2}\\ &=\frac{\sin x\cos x+\cos^2x-\sin x\cos x+\sin^2x}{(\sin x+\cos x)^2}\\ &=\bm{\frac{1}{(\sin x+\cos x)^2}} \end{aligned}

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