derivative arctan(9t)

Lösung

ableitung von

arctan (9 t)

\frac{9}{81 t ^{2} + 1}

Schritte zur Lösung

\frac{d}{d t} (arctan (9 t))

Wende die Kettenregel an:

\frac{1}{( 9 t ) ^{2} + 1} \frac{d}{d t} (9 t)

= \frac{1}{( 9 t ) ^{2} + 1} \frac{d}{d t} (9 t)

\frac{d}{d t} (9 t) = 9

= \frac{1}{( 9 t ) ^{2} + 1} \cdot 9

Vereinfache

\frac{1}{( 9 t ) ^{2} + 1} \cdot 9 : \frac{9}{81 t ^{2} + 1}

= \frac{9}{81 t ^{2} + 1}

\int_{0}^{2} x^{2} e^{- 9 x} d x

\frac{d}{d x} (6 (sec (x) - tan (x))^{\frac{3}{2}})

a re a x^{2} - 11, 10 x, [- 11, 0] y = 0

\frac{d}{d x} (- 1 0^{3 x^{2} - 4})

\frac{d}{d x} (\frac{2}{4 x ^{3}})