(1/(x+1))^'

Lösung

(\frac{1}{x + 1})^{^{'}}

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

Schritte zur Lösung

(\frac{1}{x + 1})^{'}

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

= ((x + 1)^{- 1})^{^{'}}

Wende die Kettenregel an:

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

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

(x + 1)^{'} = 1

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

Vereinfache

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

\int (2 x - cos (6 x)) d x

x \to \infty lim (0. 8^{x})

\int 3 x^{2} 2 x^{3} - 5 d x

t an g e n t f (x) = 4 - x^{2}, at x = 1.154

t an g e n t x^{3} - 3 x^{2} + 2 x - 2, at x = 0