derivative f(x)=(x^5)/5-(4x^3)/3+4x

Lösung

ableitung von

f (x) = \frac{x ^{5}}{5} - \frac{4 x ^{3}}{3} + 4 x

x^{4} - 4 x^{2} + 4

Schritte zur Lösung

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

Wende die Summen-/Differenzregel an:

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

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

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

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

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

= x^{4} - 4 x^{2} + 4

t an g e n t f (x) = 7 e^{- 0.03 x}, at x = 9

d er i v a t i v e y = \frac{( 3 x - 9 )}{( 2 x + 8 )}

\int (x^{3} + x) e^{- 4 x} d x

\int \frac{2 x}{x ^{2} - 1} d x

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