f^'(x)=(32x^3)/(2x^4+5)

Lösung

f^{^{'}} (x) = \frac{32 x ^{3}}{2 x ^{4} + 5}

f (x) = 4 ln (2 x^{4} + 5) + c_{1}

Schritte zur Lösung

f^{'} (x) = \frac{32 x ^{3}}{2 x ^{4} + 5}

Wenn

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

dann

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

f (x) = \int \frac{32 x ^{3}}{2 x ^{4} + 5} d x

\int \frac{32 x ^{3}}{2 x ^{4} + 5} d x = 4 ln (2 x^{4} + 5) + c_{1}

f (x) = 4 ln (2 x^{4} + 5) + c_{1}

f^{^{'}} (x) = 3 x^{2} + 3 x - 2.25

y^{^{'}} = - 5 \frac{x}{2} - \frac{2}{5} x

\frac{x ( 1 - x ) y ^{^{''}} - 5}{2 y ^{^{'}} - y} = 0

y^{^{''}} - 4 y + 4 y = 2 e^{2} x

f^{^{'}} (x) = \frac{2 cos ^{2} ( x )}{a ^{9}}