数式を表示したい

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三角関数の微分

• 次の関数の微分をしなさい

  $y=\sin x{\cos }^{2}x$

  $y=\frac{\sin x}{1-\cos x}$

• こたえ

  $y\text{'}=\left(\sin x\cos 2x\right)\text{'}=\left(\sin x\right)\text{'}\cos 2x+\sin x\left(\cos 2x\right)\text{'}=\cos 3x+\sin x\left(2\cos x\left(\cos x\right)\text{'}\right)=\cos 3x-2\sin 2x\cos x$

  $y\text{'}=\left(\frac{\sin x}{1-\cos x}\right)\text{'}=\frac{\left(\sin x\right)\text{'}\left(1-\cos x\right)-\sin x\left(1-\cos x\right)\text{'}}{{\left(1-\cos x\right)}^2}=\frac{\cos x\left(1-\cos x\right)-{\sin }^{2}x}{{\left(1-\cos x\right)}^2}=\frac{\cos x-\left({\cos }^{2}x+{\sin }^{2}x\right)}{{\left(1-\cos x\right)}^2}=\frac{\cos x-1}{{\left(1-\cos x\right)}^2}=-\frac{1}{1-\cos x}$

• 積分公式の例

  ${\int }_{\alpha }^{\beta }\left(x-\alpha \right)\left(\beta -x\right)dx=\frac{{\left(\beta -\alpha \right)}^3}{6}$