derivative (x^3-x)(x^2+2)(x^2+x-1)

Lösung

ableitung von

(x^{3} - x) (x^{2} + 2) (x^{2} + x - 1)

7 x^{6} + 6 x^{5} + 4 x^{3} - 9 x^{2} - 4 x + 2

Schritte zur Lösung

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

Wende die Produktregel an:

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

f = x^{3} - x, g = (x^{2} + 2) (x^{2} + x - 1)

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

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

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

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

Vereinfache

(3 x^{2} - 1) (x^{2} + 2) (x^{2} + x - 1) + (4 x^{3} + 3 x^{2} + 2 x + 2) (x^{3} - x) : 7 x^{6} + 6 x^{5} + 4 x^{3} - 9 x^{2} - 4 x + 2

= 7 x^{6} + 6 x^{5} + 4 x^{3} - 9 x^{2} - 4 x + 2

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

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

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

t an g e n t f (x) = (\frac{x + 1}{x - 3})^{2}, at x = 7

\frac{\partial}{\partial x} (x^{2} y^{2} cos (x))