What is the derivative of ln(e^{rx}+s) ?

The derivative of ln(e^{rx}+s) is (e^{rx}x)/(e^{rx)+s}

What is the first derivative of ln(e^{rx}+s) ?

The first derivative of ln(e^{rx}+s) is (e^{rx}x)/(e^{rx)+s}

derivative of

ln (e^{r x} + s)

\frac{e ^{r x} x}{e ^{r x} + s}

Solution steps

\frac{d}{d r} (ln (e^{r x} + s))

Treat

x, s

as constants

Apply the chain rule:

\frac{1}{e ^{r x} + s} \frac{d}{d r} (e^{r x} + s)

= \frac{1}{e ^{r x} + s} \frac{d}{d r} (e^{r x} + s)

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

= \frac{1}{e ^{r x} + s} x e^{x r}

Simplify

\frac{1}{e ^{r x} + s} x e^{x r} : \frac{e ^{r x} x}{e ^{r x} + s}

= \frac{e ^{r x} x}{e ^{r x} + s}

d er i v a t i v e f (x) = x^{2} - 2 x - 2

\frac{\partial}{\partial x} (1 0^{- 2 x})

\int a rcco tan (x) d x

\frac{d}{d x} (6 x^{3} x csc (x))

a re a x^{2}, x + 2, - 1, 2

What is the derivative of ln(e^{rx}+s) ?
The derivative of ln(e^{rx}+s) is (e^{rx}x)/(e^{rx)+s}
What is the first derivative of ln(e^{rx}+s) ?
The first derivative of ln(e^{rx}+s) is (e^{rx}x)/(e^{rx)+s}