f^'(x)=2x

Lösung

f^{^{'}} (x) = 2 x

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

Schritte zur Lösung

f^{'} (x) = 2 x

Wenn

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

dann

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

f (x) = \int 2 x d x

\int 2 x d x = x^{2} + c_{1}

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

\int cos^{2} (\frac{4 - x}{3}) d x

\frac{d}{d t} (\frac{1}{2} t)

\int x^{2} (x^{3} + 9)^{7} d x

d er i v a t i v e f (x) = x^{\frac{7}{5}} (x + 5)

y^{^{'}} = (0.04 \cdot 4) - \frac{y}{1840} 4, y (0) = 0