derivative of \sqrt[3]{x^2}(2x-x^2)

Lösung

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

\frac{2 x ^{\frac{2}{3}} ( - x + 2 )}{3} + 3 x^{2} (2 - 2 x)

Schritte zur Lösung

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 3 x^{2}, g = 2 x - x^{2}

= \frac{d}{d x} (3 x^{2}) (2 x - x^{2}) + \frac{d}{d x} (2 x - x^{2}) 3 x^{2}

\frac{d}{d x} (3 x^{2}) = \frac{2}{3 x ^{\frac{1}{3}}}

\frac{d}{d x} (2 x - x^{2}) = 2 - 2 x

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

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

= \frac{2 x ^{\frac{2}{3}} ( - x + 2 )}{3} + 3 x^{2} (2 - 2 x)

\frac{d}{d t} (a e^{b t})

\frac{\partial}{\partial x} (e^{- t})

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

\frac{\partial}{\partial t} (sech^{2} (x - t))

\int 6 x^{4} e^{x^{5}} d x