f^'(x)=2^x

Lösung

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

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

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

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

x \to 0 lim (ln (x) x)

\frac{\partial}{\partial x} (x^{2} y^{3} + 6 x yz - 2 yz)

\int \frac{1}{1 + sin ^{2} ( x )} d x

\frac{\partial}{\partial x} (7 x e^{x^{2}} + y^{2})