derivative f(x)=(x^3-10x)(3x^2+4x)

Lösung

ableitung von

f (x) = (x^{3} - 10 x) (3 x^{2} + 4 x)

15 x^{4} + 16 x^{3} - 90 x^{2} - 80 x

Schritte zur Lösung

\frac{d}{d x} ((x^{3} - 10 x) (3 x^{2} + 4 x))

Wende die Produktregel an:

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

f = x^{3} - 10 x, g = 3 x^{2} + 4 x

= \frac{d}{d x} (x^{3} - 10 x) (3 x^{2} + 4 x) + \frac{d}{d x} (3 x^{2} + 4 x) (x^{3} - 10 x)

\frac{d}{d x} (x^{3} - 10 x) = 3 x^{2} - 10

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

= (3 x^{2} - 10) (3 x^{2} + 4 x) + (6 x + 4) (x^{3} - 10 x)

Vereinfache

(3 x^{2} - 10) (3 x^{2} + 4 x) + (6 x + 4) (x^{3} - 10 x) : 15 x^{4} + 16 x^{3} - 90 x^{2} - 80 x

= 15 x^{4} + 16 x^{3} - 90 x^{2} - 80 x

a re a y = x^{2} ln (x), y = 4 ln (x)

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

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

\int_{- 8}^{8} (4 x^{9} + 7 x^{5}) d x

t an g e n t y = \frac{( x + 5 ) ^{5}}{7 x - 5}