d/(d{r)}(e^{{r}}+{r}^e)

Lösung

\frac{d}{d r} (e^{r} + r^{e})

e^{r} + e r^{e - 1}

Schritte zur Lösung

\frac{d}{d r} (e^{r} + r^{e})

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d r} (e^{r}) + \frac{d}{d r} (r^{e})

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

\frac{d}{d r} (r^{e}) = e r^{e - 1}

= e^{r} + e r^{e - 1}

a re a 13 sin (x), 13 cos (x), 0, π

x \to \infty lim ([0, x])

t \to 0 - lim (2 t - 4)

x \to 0 - lim (- 2 + \frac{18}{x ^{2}})

x \to 6 + lim (\frac{3}{x - 6})