What is the d/(d{r)}(e^{{r}}+{r}^e) ?

The d/(d{r)}(e^{{r}}+{r}^e) is e^{{r}}+e{r}^{e-1}

What is the first d/(d{r)}(e^{{r}}+{r}^e) ?

The first d/(d{r)}(e^{{r}}+{r}^e) is e^{{r}}+e{r}^{e-1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(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}

\int cot (24 x) d x

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}})

What is the d/(d{r)}(e^{{r}}+{r}^e) ?
The d/(d{r)}(e^{{r}}+{r}^e) is e^{{r}}+e{r}^{e-1}
What is the first d/(d{r)}(e^{{r}}+{r}^e) ?
The first d/(d{r)}(e^{{r}}+{r}^e) is e^{{r}}+e{r}^{e-1}