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

Lösung

ableitung von

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

18 x^{5} + 10 x^{4} - 44 x^{3} - 3 x^{2} + 24 x - 5

Schritte zur Lösung

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

Wende die Produktregel an:

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

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

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

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

\frac{d}{d x} (3 x^{3} + 2 x^{2} - 5 x) = 9 x^{2} + 4 x - 5

= (3 x^{2} - 2) (3 x^{3} + 2 x^{2} - 5 x) + (9 x^{2} + 4 x - 5) (x^{3} - 2 x + 1)

Vereinfache

(3 x^{2} - 2) (3 x^{3} + 2 x^{2} - 5 x) + (9 x^{2} + 4 x - 5) (x^{3} - 2 x + 1) : 18 x^{5} + 10 x^{4} - 44 x^{3} - 3 x^{2} + 24 x - 5

= 18 x^{5} + 10 x^{4} - 44 x^{3} - 3 x^{2} + 24 x - 5

d er i v a t i v e y = \frac{( x + 1 ) ^{2}}{x + 1}

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} d y + \frac{t}{2 y ^{4}} d t = 0

n = 0 \sum \infty (\frac{3}{5})^{n}

x \to + 3 - lim (\frac{3}{x - 3})

\int x \cdot 2^{x} d x