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

Lösung

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

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

Schritte zur Lösung

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

Behandle

t

als Konstante

Wende die Summen-/Differenzregel an:

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

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

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

\frac{d}{d x} (2 t^{3}) = 0

= \frac{1}{3 x ^{\frac{2}{3}}} + 0

Vereinfache

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

\frac{\partial}{\partial x} (l sin (θ))

d er i v a t i v e \frac{( cos ^{3} ( x ) - 1 )}{cos ^{2} ( x )}

11 (t + 1) y^{^{'}} - 9 y = 18 t

\frac{d}{d x} (\frac{x}{x + \frac{b}{x}})

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