derivative of 4^x+1

Lösung

\frac{d}{d x} (4^{x} + 1)

ln (2) \cdot 2^{2 x + 1}

Schritte zur Lösung

\frac{d}{d x} (4^{x} + 1)

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (4^{x}) + \frac{d}{d x} (1)

\frac{d}{d x} (4^{x}) = ln (2) \cdot 2^{2 x + 1}

\frac{d}{d x} (1) = 0

= ln (2) \cdot 2^{2 x + 1} + 0

Vereinfache

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

\frac{d}{d x} (x^{2} + 3 x + 5)

x \to 2 lim (\frac{4}{( x - 2 ) ^{3}})

\frac{d}{d x} (d cos (x) + x^{2} sin (x))

n = 1 \sum \infty \frac{3}{n ^{\frac{5}{3}}}

s l o p e ∣ x ∣ + 3