(ln(x^2+1))^'

Lösung

(ln (x^{2} + 1))^{^{'}}

\frac{2 x}{x ^{2} + 1}

Schritte zur Lösung

(ln (x^{2} + 1))^{'}

Wende die Kettenregel an:

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

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

(x^{2} + 1)^{'} = 2 x

= \frac{1}{x ^{2} + 1} \cdot 2 x

Vereinfache

\frac{1}{x ^{2} + 1} \cdot 2 x : \frac{2 x}{x ^{2} + 1}

= \frac{2 x}{x ^{2} + 1}

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (- 1) = 0, y^{^{'}} (- 1) = 1

\int \frac{3 x + 5}{x ( 2 x - 1 )} d x

d er i v a t i v e t^{3} e^{t} t

n \to \infty lim (c^{n})

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 1, y^{^{'}} (0) = 1