derivative (2x+8x^2)/((1+8x)^2)

Lösung

ableitung von

\frac{2 x + 8 x ^{2}}{( 1 + 8 x ) ^{2}}

\frac{2}{( 1 + 8 x ) ^{3}}

Schritte zur Lösung

\frac{d}{d x} (\frac{2 x + 8 x ^{2}}{( 1 + 8 x ) ^{2}})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 2 x + 8 x ^{2} ) ( 1 + 8 x ) ^{2} - \frac{d}{d x} ( ( 1 + 8 x ) ^{2} ) ( 2 x + 8 x ^{2} )}{( ( 1 + 8 x ) ^{2} ) ^{2}}

\frac{d}{d x} (2 x + 8 x^{2}) = 2 + 16 x

\frac{d}{d x} ((1 + 8 x)^{2}) = 16 (1 + 8 x)

= \frac{( 2 + 16 x ) ( 1 + 8 x ) ^{2} - 16 ( 1 + 8 x ) ( 2 x + 8 x ^{2} )}{( ( 1 + 8 x ) ^{2} ) ^{2}}

Vereinfache

\frac{( 2 + 16 x ) ( 1 + 8 x ) ^{2} - 16 ( 1 + 8 x ) ( 2 x + 8 x ^{2} )}{( ( 1 + 8 x ) ^{2} ) ^{2}} : \frac{2}{( 1 + 8 x ) ^{3}}

= \frac{2}{( 1 + 8 x ) ^{3}}

\int_{0}^{\frac{1}{3}} \frac{4}{1 + 9 x ^{2}} d x

t an g e n t y = x^{3} - 2 x^{2} + 5

\int r ln (r) d r

x \to 1 lim (\frac{x ^{8} - 1}{x - 1})

n = 1 \sum \infty \frac{( 3 ^{n - 1} )}{9 ^{n}}