derivative (1+sqrt(t))(8t^2-1)

Lösung

ableitung von

(1 + t) (8 t^{2} - 1)

\frac{40 t ^{2} + 32 t t - 1}{2 t}

Schritte zur Lösung

\frac{d}{d t} ((1 + t) (8 t^{2} - 1))

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 1 + t, g = 8 t^{2} - 1

= \frac{d}{d t} (1 + t) (8 t^{2} - 1) + \frac{d}{d t} (8 t^{2} - 1) (1 + t)

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

\frac{d}{d t} (8 t^{2} - 1) = 16 t

= \frac{1}{2 t} (8 t^{2} - 1) + 16 t (1 + t)

Vereinfache

\frac{1}{2 t} (8 t^{2} - 1) + 16 t (1 + t) : \frac{40 t ^{2} + 32 t t - 1}{2 t}

= \frac{40 t ^{2} + 32 t t - 1}{2 t}

x \to \infty lim (3 x^{4})

x \to 0 lim (\frac{e ^{x} - 1}{7 x})

\frac{\partial}{\partial x} (- 5 e^{- 5 y} sin (5 x))

x \to 2 lim (2 - x)

\frac{d}{d x} (2 e^{c o s (\frac{x}{2})})