derivative of e^x+2^{-x}+2cos(x-6)

Lösung

\frac{d}{d x} (e^{x} + 2^{- x} + 2 cos (x) - 6)

e^{x} - ln (2) \cdot 2^{- x} - 2 sin (x)

Schritte zur Lösung

\frac{d}{d x} (e^{x} + 2^{- x} + 2 cos (x) - 6)

Wende die Summen-/Differenzregel an:

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

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

\frac{d}{d x} (e^{x}) = e^{x}

\frac{d}{d x} (2^{- x}) = - ln (2) \cdot 2^{- x}

\frac{d}{d x} (2 cos (x)) = - 2 sin (x)

\frac{d}{d x} (6) = 0

= e^{x} - ln (2) \cdot 2^{- x} - 2 sin (x) - 0

Vereinfache

= e^{x} - ln (2) \cdot 2^{- x} - 2 sin (x)

\int \frac{1}{12 x ^{2}} d x

\int \frac{3 e ^{x}}{e ^{2 x} + 2 e ^{x} + 1} d x

\int_{0}^{2 π} e^{i θ} d θ

\int_{0}^{\infty} 4 x e^{- 4 x} d x

d er i v a t i v e cos (sin (7 x))