d/(dt)(atcos(4t)+btsin(4t))

Lösung

\frac{d}{d t} (a t cos (4 t) + b t sin (4 t))

a (cos (4 t) - 4 t sin (4 t)) + b (sin (4 t) + 4 t cos (4 t))

Schritte zur Lösung

\frac{d}{d t} (a t cos (4 t) + b t sin (4 t))

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (a t cos (4 t)) + \frac{d}{d t} (b t sin (4 t))

\frac{d}{d t} (a t cos (4 t)) = a (cos (4 t) - 4 t sin (4 t))

\frac{d}{d t} (b t sin (4 t)) = b (sin (4 t) + 4 t cos (4 t))

= a (cos (4 t) - 4 t sin (4 t)) + b (sin (4 t) + 4 t cos (4 t))

\frac{\partial}{\partial x} (- x^{2} sin (x^{2} y))

n \to \infty lim (\frac{4}{n})

\frac{d}{d x} ((2 x + e^{x}) (5 - x))

p a r t ia l f r a c t i o n \frac{2 x ^{2}}{x ^{2} - 1}

\frac{\partial}{\partial x} (e^{x^{2} y^{4}} sin (x yz))