derivative of (sin(x+1)/(e^x))

解

\frac{d}{d x} (\frac{sin ( x + 1 )}{e ^{x}})

\frac{cos ( 1 + x ) - sin ( 1 + x )}{e ^{x}}

解答ステップ

\frac{d}{d x} (\frac{sin ( x + 1 )}{e ^{x}})

商の法則を適用:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( sin ( x + 1 ) ) e ^{x} - \frac{d}{d x} ( e ^{x} ) sin ( x + 1 )}{( e ^{x} ) ^{2}}

\frac{d}{d x} (sin (x + 1)) = cos (x + 1)

\frac{d}{d x} (e^{x}) = e^{x}

= \frac{cos ( x + 1 ) e ^{x} - e ^{x} sin ( x + 1 )}{( e ^{x} ) ^{2}}

簡素化

\frac{cos ( x + 1 ) e ^{x} - e ^{x} sin ( x + 1 )}{( e ^{x} ) ^{2}} : \frac{cos ( 1 + x ) - sin ( 1 + x )}{e ^{x}}

= \frac{cos ( 1 + x ) - sin ( 1 + x )}{e ^{x}}