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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\frac{d}{d t} (\frac{1}{t ^{2} + 1})

t an g e n t f (x) = x + 1 - ln (x), at x = 0.5

(t^{2} + x^{2}) d t + (t x) d x = 0

\frac{d}{d x} (ln (x x^{2} - 7))

in v erse l a pl a ce \frac{1}{s ^{2} \cdot ( s ^{2} + 4 )}