d/(dt)(4t^2+2t+5)

Lösung

\frac{d}{d t} (4 t^{2} + 2 t + 5)

8 t + 2

Schritte zur Lösung

\frac{d}{d t} (4 t^{2} + 2 t + 5)

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d t} (4 t^{2}) + \frac{d}{d t} (2 t) + \frac{d}{d t} (5)

\frac{d}{d t} (4 t^{2}) = 8 t

\frac{d}{d t} (2 t) = 2

\frac{d}{d t} (5) = 0

= 8 t + 2 + 0

Vereinfache

= 8 t + 2

\frac{d}{d x} (e^{c o s (x)} + e^{s i n (y)})

\frac{d}{d x} (lo g_{3} (5 x - 1))

d er i v a t i v e \frac{1}{2 r} + \frac{1}{9 r ^{\frac{8}{9}}}

\int - 9.8 d x

\frac{\partial}{\partial x} (11 x^{2} y + 8 x y^{3})