(1/(1-x))^'

Lösung

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

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

Schritte zur Lösung

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

Wende Exponentenregel an:

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

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

l a pl a ce t r an s f or m e^{2 t} sin (j^{2} t)

\frac{\partial}{\partial x} ((9 + 20 x) y^{t a n (x)})

t an g e n t f (x) = - x^{2}, (- 3, - 9)

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

x \to - \infty lim (- 3 - x)