derivative of (3x^2+2x-1/((1+3x^2)^2))

Lösung

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

\frac{2 ( - 9 x ^{3} - 9 x ^{2} + 9 x + 1 )}{( 1 + 3 x ^{2} ) ^{3}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (3 x^{2} + 2 x - 1) = 6 x + 2

\frac{d}{d x} ((1 + 3 x^{2})^{2}) = 12 x (1 + 3 x^{2})

= \frac{( 6 x + 2 ) ( 1 + 3 x ^{2} ) ^{2} - 12 x ( 1 + 3 x ^{2} ) ( 3 x ^{2} + 2 x - 1 )}{( ( 1 + 3 x ^{2} ) ^{2} ) ^{2}}

Vereinfache

\frac{( 6 x + 2 ) ( 1 + 3 x ^{2} ) ^{2} - 12 x ( 1 + 3 x ^{2} ) ( 3 x ^{2} + 2 x - 1 )}{( ( 1 + 3 x ^{2} ) ^{2} ) ^{2}} : \frac{2 ( - 9 x ^{3} - 9 x ^{2} + 9 x + 1 )}{( 1 + 3 x ^{2} ) ^{3}}

= \frac{2 ( - 9 x ^{3} - 9 x ^{2} + 9 x + 1 )}{( 1 + 3 x ^{2} ) ^{3}}

y^{^{'}} = 2 (t + 1) y, y (0) = 1

y^{^{''}} + 2 \cdot y^{^{'}} + y = e^{- t}

n = 0 \sum \infty \frac{1}{2 ^{n} n !}

d er i v a t i v e f (x) = \frac{( x ^{3} + 1 )}{( x ^{4} + 1 )}

x \to - 1 lim (\frac{x + 1}{x ^{2} - 1})