derivative f(x)=(3x^3)^{0.5}

Lösung

ableitung von

f (x) = (3 x^{3})^{0.5}

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

Schritte zur Lösung

\frac{d}{d x} ((3 x^{3})^{0.5})

Vereinfache

(3 x^{3})^{0.5} : 3^{\frac{1}{2}} (x^{3})^{\frac{1}{2}}

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

Entferne die Konstante:

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

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

Wende die Potenzregel an:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

Vereinfache

3^{\frac{1}{2}} \cdot \frac{1}{2} (x^{3})^{\frac{1}{2} - 1} : \frac{3 ^{\frac{1}{2}}}{2 x ^{\frac{1}{2}} ( x ^{2} ) ^{\frac{1}{2}}}

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

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

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

d er i v a t i v e - 6 s

x \to - 3 lim (\frac{1}{x ^{2} - 9})

t a y l or \frac{1}{2 + x}, 0