derivative of 2ln(1+x^2)

Lösung

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

\frac{4 x}{1 + x ^{2}}

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

2 \cdot \frac{1}{1 + x ^{2}} \cdot 2 x : \frac{4 x}{1 + x ^{2}}

= \frac{4 x}{1 + x ^{2}}

x \to \infty lim (\frac{7 x - 1}{3 5 x ^{3} + 4 x - 1})

d er i v a t i v e \frac{1}{( x ^{2} + 8 ) ^{5}}

\int p (p + 6)^{6} d p

\int 5 \frac{x}{16 - x ^{2}} d x

\int (- 5 x^{4} + 8 x^{3}) (- 2 x^{5} + 4 x^{4})^{6} d x