微分の計算問題

${\displaystyle y^{\prime }=\frac{{\left(1-3sinx\right)}^{\prime }2cosx-\left(1-3sinx\right){\left(2cosx\right)}^{\prime }}{{\left(2cosx\right)}^2}}$

( 分数関数の微分を参照)

${\displaystyle y^{\prime }=\frac{{\left(1-3\sin x\right)}^{\prime }2\cos x-\left(1-3\sin x\right){\left(2\cos x\right)}^{\prime }}{{\left(2\cos x\right)}^2}}$

${\displaystyle =\frac{-3\cos x·2\cos x-\left(1-3\sin x\right)\left(-2\sin x\right)}{{\left(2\cos x\right)}^2}}$

${\displaystyle =\frac{-6{\cos }^{2}x-6{\sin }^{2}x+2\sin x}{{\left(2\cos x\right)}^2}}$

${\displaystyle =\frac{-6\left({sin}^{2}x+{cos}^{2}x\right)+2sinx}{{\left(2cosx\right)}^2}}$

(三角関数の相互関係の関係式より, ${\displaystyle {sin}^{2}x+{cos}^{2}x=1}$ )

${\displaystyle =\frac{2\left(sinx-3\right)}{{\left(2cosx\right)}^2}}$

${\displaystyle =\frac{sinx-3}{2{cos}^{2}x}}$