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

Lösung

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

6 x^{5} + 5 x^{4} - 8 x^{3} - 6 x^{2} + 2 x

Schritte zur Lösung

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

Wende die Produktregel an:

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

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

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

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

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

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

Vereinfache

(4 x^{3} + 3 x^{2}) (x^{2} - 2) + 2 x (x^{4} + x^{3} + 1) : 6 x^{5} + 5 x^{4} - 8 x^{3} - 6 x^{2} + 2 x

= 6 x^{5} + 5 x^{4} - 8 x^{3} - 6 x^{2} + 2 x

\int \frac{36 + x ^{2}}{x} d x

\int_{- 9}^{- 5} \frac{1}{x} d x

\int_{0}^{5} (6 e^{0.2 x} + 2 x) d x

d er i v a t i v e f (x) = \frac{5}{x ^{3}} - \frac{4}{x ^{2}} + \frac{1}{x}

\frac{d}{d x} (2 tan (x) - 4 sec (x))