derivative of 2^{3x-1}

Lösung

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

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

Schritte zur Lösung

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

Wende Exponentenregel an:

a^{b} = e^{b l n (a)}

2^{3 x - 1} = e^{(3 x - 1) l n (2)}

= \frac{d}{d x} (e^{(3 x - 1) l n (2)})

Wende die Kettenregel an:

e^{(3 x - 1) l n (2)} \frac{d}{d x} ((3 x - 1) ln (2))

= e^{(3 x - 1) l n (2)} \frac{d}{d x} ((3 x - 1) ln (2))

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

= e^{(3 x - 1) l n (2)} \cdot 3 ln (2)

e^{(3 x - 1) l n (2)} = 2^{3 x - 1}

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

t an g e n t y = \frac{x}{1 + x ^{2}}, (3, 0.3)

x \to 0 + lim (\frac{2}{x} - \frac{2}{∣ x ∣})

\frac{d}{d x} (c x e^{3 x})

\frac{d}{d t} (sin (t) + t cos (t))

\frac{d}{d x} (ln ((6 x + 4) (3 x^{2} + 2)))