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

The d/(dt)(e^{{r}(t)t}) is e^{{r}(t)t}(t d/(dt)({r}(t))+{r}(t))

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

The first d/(dt)(e^{{r}(t)t}) is e^{{r}(t)t}(t d/(dt)({r}(t))+{r}(t))

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d t} (r (t) t) = t \frac{d}{d t} (r (t)) + r (t)

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

- x^{2} + 9 x, x^{2} - 5 x

y = 3 - 5 x

\int_{4}^{9} \frac{1}{8 - x} d x

\int e^{- x t} d x

\frac{d}{d x} (3 ln (x) + ln (y))

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