derivative of 1/9 (x^2+6^{3/2})

Lösung

\frac{d}{d x} (\frac{1}{9} (x^{2} + 6)^{\frac{3}{2}})

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

Schritte zur Lösung

\frac{d}{d x} (\frac{1}{9} (x^{2} + 6)^{\frac{3}{2}})

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{9} \frac{d}{d x} ((x^{2} + 6)^{\frac{3}{2}})

Wende die Kettenregel an:

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

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

\frac{d}{d x} (x^{2} + 6) = 2 x

= \frac{1}{9} \cdot \frac{3 ( x ^{2} + 6 ) ^{\frac{1}{2}}}{2} \cdot 2 x

Vereinfache

\frac{1}{9} \cdot \frac{3 ( x ^{2} + 6 ) ^{\frac{1}{2}}}{2} \cdot 2 x : \frac{x ( x ^{2} + 6 ) ^{\frac{1}{2}}}{3}

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

y^{^{'}} + \frac{1}{x} y = 9 x^{2}

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

a re a 3 x, x

\int \frac{sec ^{2} ( 3 x )}{1 + tan ( 3 x )} d x

s l o p e y = x^{2} - 3 x - 5, (2, - 7)