derivative ln(t^2+1)

Lösung

ableitung von

ln (t^{2} + 1)

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

x \to 0 lim (\frac{x ^{3}}{x})

\int \frac{2 x ^{4} + 7 x ^{3} - x}{x ^{3}} d x

\frac{\partial}{\partial t} (x e^{2 t} - 4 x)

\int_{- 1}^{2} 2^{x} d x

n = 0 \sum \infty e^{- \frac{n}{2}}