derivative (1+sqrt(t))(2t^2-5)

Lösung

ableitung von

(1 + t) (2 t^{2} - 5)

\frac{10 t ^{2} + 8 t t - 5}{2 t}

Schritte zur Lösung

\frac{d}{d t} ((1 + t) (2 t^{2} - 5))

Wende die Produktregel an:

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

f = 1 + t, g = 2 t^{2} - 5

= \frac{d}{d t} (1 + t) (2 t^{2} - 5) + \frac{d}{d t} (2 t^{2} - 5) (1 + t)

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

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

= \frac{1}{2 t} (2 t^{2} - 5) + 4 t (1 + t)

Vereinfache

\frac{1}{2 t} (2 t^{2} - 5) + 4 t (1 + t) : \frac{10 t ^{2} + 8 t t - 5}{2 t}

= \frac{10 t ^{2} + 8 t t - 5}{2 t}

d er i v a t i v e f (x) = (\frac{4 x ^{2}}{2 - x})^{3}

s l o p e \frac{11 - 3}{p - 1}

\frac{\partial}{\partial x} (y^{\frac{x}{z}})

d er i v a t i v e f (x) = (2 x - x) (5 x + 2)

\int \frac{- 2 x}{( 1 + x ^{2} ) ^{2}} d x