f^'(x)=4x+2

Lösung

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

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

Schritte zur Lösung

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

Wenn

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

dann

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

f (x) = \int 4 x + 2 d x

\int 4 x + 2 d x = 2 x^{2} + 2 x + c_{1}

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

\int x^{2} x^{2} + 3 d x

\frac{\partial}{\partial x} (\frac{5 x}{( 3 + y )})

y^{^{''}} - 2 y^{^{'}} = 2 x

d er i v a t i v e f (x) = x^{- \frac{2}{5}} + x^{\frac{2}{5}}

\frac{\partial}{\partial y} (\frac{y ^{2}}{1 + x ^{2}})