derivative of (e^{x^2}/((2x+1)^5))

解

\frac{d}{d x} (\frac{e ^{x^{2}}}{( 2 x + 1 ) ^{5}})

\frac{2 e ^{x^{2}} ( x ( 2 x + 1 ) - 5 )}{( 2 x + 1 ) ^{6}}

解答ステップ

\frac{d}{d x} (\frac{e ^{x^{2}}}{( 2 x + 1 ) ^{5}})

商の法則を適用:

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

= \frac{\frac{d}{d x} ( e ^{x^{2}} ) ( 2 x + 1 ) ^{5} - \frac{d}{d x} ( ( 2 x + 1 ) ^{5} ) e ^{x^{2}}}{( ( 2 x + 1 ) ^{5} ) ^{2}}

\frac{d}{d x} (e^{x^{2}}) = e^{x^{2}} \cdot 2 x

\frac{d}{d x} ((2 x + 1)^{5}) = 10 (2 x + 1)^{4}

= \frac{e ^{x^{2}} \cdot 2 x ( 2 x + 1 ) ^{5} - 10 ( 2 x + 1 ) ^{4} e ^{x^{2}}}{( ( 2 x + 1 ) ^{5} ) ^{2}}

簡素化

\frac{e ^{x^{2}} \cdot 2 x ( 2 x + 1 ) ^{5} - 10 ( 2 x + 1 ) ^{4} e ^{x^{2}}}{( ( 2 x + 1 ) ^{5} ) ^{2}} : \frac{2 e ^{x^{2}} ( x ( 2 x + 1 ) - 5 )}{( 2 x + 1 ) ^{6}}

= \frac{2 e ^{x^{2}} ( x ( 2 x + 1 ) - 5 )}{( 2 x + 1 ) ^{6}}