derivative of 4+2x-3x^2-5x^3-8x^4+9x^5

Lösung

\frac{d}{d x} (4 + 2 x - 3 x^{2} - 5 x^{3} - 8 x^{4} + 9 x^{5})

45 x^{4} - 32 x^{3} - 15 x^{2} - 6 x + 2

Schritte zur Lösung

\frac{d}{d x} (4 + 2 x - 3 x^{2} - 5 x^{3} - 8 x^{4} + 9 x^{5})

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (4) + \frac{d}{d x} (2 x) - \frac{d}{d x} (3 x^{2}) - \frac{d}{d x} (5 x^{3}) - \frac{d}{d x} (8 x^{4}) + \frac{d}{d x} (9 x^{5})

\frac{d}{d x} (4) = 0

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

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

\frac{d}{d x} (5 x^{3}) = 15 x^{2}

\frac{d}{d x} (8 x^{4}) = 32 x^{3}

\frac{d}{d x} (9 x^{5}) = 45 x^{4}

= 0 + 2 - 6 x - 15 x^{2} - 32 x^{3} + 45 x^{4}

Vereinfache

= 45 x^{4} - 32 x^{3} - 15 x^{2} - 6 x + 2

\int_{0}^{2 π} 20 + 16 t^{2} d t

x \to - 2 lim ((x + 1)^{4} (2 x^{2}))

d er i v a t i v e \frac{x ^{2} + 9}{x + 3}

\int_{0}^{2} \frac{1}{4 + x ^{2}} d x

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