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

Lösung

\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}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

(\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}}

Vereinfache

\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}}

\frac{\partial}{\partial y} (4 x y^{3})

n = 1 \sum \infty \frac{2}{n \cdot ( n + 2 )}

f (x) = 2 sec (x)

\frac{d}{d x} (80 (1 - 0.1 x)^{3})

\frac{d}{d x} (ln (\frac{x + b}{x}))