derivative f(t)=e^{9tsin(2t)}

Lösung

ableitung von

f (t) = e^{9 t s i n (2 t)}

e^{9 t s i n (2 t)} \cdot 9 (sin (2 t) + 2 t cos (2 t))

Schritte zur Lösung

\frac{d}{d t} (e^{9 t s i n (2 t)})

Wende die Kettenregel an:

e^{9 t s i n (2 t)} \frac{d}{d t} (9 t sin (2 t))

= e^{9 t s i n (2 t)} \frac{d}{d t} (9 t sin (2 t))

\frac{d}{d t} (9 t sin (2 t)) = 9 (sin (2 t) + 2 t cos (2 t))

= e^{9 t s i n (2 t)} \cdot 9 (sin (2 t) + 2 t cos (2 t))

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

\int 3 sec^{- 2} (x t a) n^{3} x d x

x \to 9 lim (2 x^{2} - 3)

\frac{\partial}{\partial x} (sin (x^{3}) + 3 x y)

l a pl a ce t r an s f or m t^{2} cos (4 t)