f^'x=(2x)(x^4-8x)+(x^2)(4x^3-8)

Lösung

f^{^{'}} x = (2 x) (x^{4} - 8 x) + (x^{2}) (4 x^{3} - 8)

f (x) = 6 (\frac{x ^{5}}{5} - 2 x^{2}) + c_{1}

Schritte zur Lösung

f^{'} (x) x = (2 x) (x^{4} - 8 x) + (x^{2}) (4 x^{3} - 8)

Isoliere

f^{'} (x) : f^{'} (x) = 6 x (x^{3} - 4)

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

f (x) = \int 6 x (x^{3} - 4) d x

\int 6 x (x^{3} - 4) d x = 6 (\frac{x ^{5}}{5} - 2 x^{2}) + c_{1}

f (x) = 6 (\frac{x ^{5}}{5} - 2 x^{2}) + c_{1}

d er i v a t i v e y = \frac{6 t - 3}{5 t + 8}

x \to \frac{π}{2} lim (sin (\frac{x}{2}))

n = 3 \sum \infty (\frac{e}{π})^{n}

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

in v erse l a pl a ce \frac{1}{( s ^{2} + 4 ) ( s ^{2} + 1 )}