((x^3)/6+1/(2x))^'

Lösung

(\frac{x ^{3}}{6} + \frac{1}{2 x})^{^{'}}

\frac{x ^{2}}{2} - \frac{1}{2 x ^{2}}

Schritte zur Lösung

(\frac{x ^{3}}{6} + \frac{1}{2 x})^{'}

Wende die Summen-/Differenzregel an:

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

= (\frac{x ^{3}}{6})^{^{'}} + (\frac{1}{2 x})^{^{'}}

(\frac{x ^{3}}{6})^{'} = \frac{x ^{2}}{2}

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

= \frac{x ^{2}}{2} - \frac{1}{2 x ^{2}}

\int_{6}^{\infty} \frac{1}{( x - 5 ) ^{\frac{3}{2}}} d x

\int_{0}^{5} x^{2} + 25 d x

d er i v a t i v e y = sin (x y)

\frac{\partial}{\partial t} (x ln (t))

\int cos^{3} (\frac{x}{10}) d x