derivative of ((x-1/(x^2+x+1))^4)

Lösung

\frac{d}{d x} ((\frac{x - 1}{x ^{2} + x + 1})^{4})

\frac{4 ( x - 1 ) ^{3} ( - x ^{2} + 2 x + 2 )}{( x ^{2} + x + 1 ) ^{5}}

Schritte zur Lösung

\frac{d}{d x} ((\frac{x - 1}{x ^{2} + x + 1})^{4})

Wende die Kettenregel an:

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

= 4 (\frac{x - 1}{x ^{2} + x + 1})^{3} \frac{d}{d x} (\frac{x - 1}{x ^{2} + x + 1})

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

= 4 (\frac{x - 1}{x ^{2} + x + 1})^{3} \frac{- x ^{2} + 2 x + 2}{( x ^{2} + x + 1 ) ^{2}}

Vereinfache

4 (\frac{x - 1}{x ^{2} + x + 1})^{3} \frac{- x ^{2} + 2 x + 2}{( x ^{2} + x + 1 ) ^{2}} : \frac{4 ( x - 1 ) ^{3} ( - x ^{2} + 2 x + 2 )}{( x ^{2} + x + 1 ) ^{5}}

= \frac{4 ( x - 1 ) ^{3} ( - x ^{2} + 2 x + 2 )}{( x ^{2} + x + 1 ) ^{5}}

\int sin (x) + 3 d x

t an g e n t f (x) = x^{2} - x, at x = 3

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

d er i v a t i v e lo g_{b} (x)

\frac{d}{d x} (2 sin (x) + cos (3 x))