(x^2+x)^'

Lösung

(x^{2} + x)^{^{'}}

2 x + 1

Schritte zur Lösung

(x^{2} + x)^{'}

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

x^{'} = 1

= 2 x + 1

\frac{d}{d x} (\frac{( x ^{2} )}{sin ( x )})

y^{^{'}} - e^{6 x + y} = 0

\int x (x + 1)^{4} d x

(arcsin (2 x))^{^{'}}

y^{^{''}} - 9 y = e^{3 x}