derivative of (2x+5(3x^2-4x+1))

Lösung

\frac{d}{d x} ((2 x + 5) (3 x^{2} - 4 x + 1))

18 x^{2} + 14 x - 18

Schritte zur Lösung

\frac{d}{d x} ((2 x + 5) (3 x^{2} - 4 x + 1))

Wende die Produktregel an:

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

f = 2 x + 5, g = 3 x^{2} - 4 x + 1

= \frac{d}{d x} (2 x + 5) (3 x^{2} - 4 x + 1) + \frac{d}{d x} (3 x^{2} - 4 x + 1) (2 x + 5)

\frac{d}{d x} (2 x + 5) = 2

\frac{d}{d x} (3 x^{2} - 4 x + 1) = 6 x - 4

= 2 (3 x^{2} - 4 x + 1) + (6 x - 4) (2 x + 5)

Vereinfache

2 (3 x^{2} - 4 x + 1) + (6 x - 4) (2 x + 5) : 18 x^{2} + 14 x - 18

= 18 x^{2} + 14 x - 18

ma c l a u r in (x + 1) (ln (x + 1))

a re a e^{2 x}, e^{5 x}, x = 1

\frac{\partial}{\partial x} (x 1 + y^{9})

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

\frac{d}{d x} (\frac{1}{1 + 4 x ^{2} + 9 x ^{4}})