derivative f(t)=2t^3+t

Lösung

ableitung von

f (t) = 2 t^{3} + t

6 t^{2} + 1

Schritte zur Lösung

\frac{d}{d t} (2 t^{3} + t)

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d t} (2 t^{3}) + \frac{d t}{d t}

\frac{d}{d t} (2 t^{3}) = 6 t^{2}

\frac{d t}{d t} = 1

= 6 t^{2} + 1

\frac{\partial}{\partial x} (\frac{x y}{x + y})

d er i v a t i v e f (x) = ln (6 - x^{2})

\frac{\partial}{\partial x} (x cos (y) - y sin (x))

x \to 1 lim (\frac{( 5 - 5 x )}{1 - x})

x \frac{d y}{d x} - y = 0, y (a) = b