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}

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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}

f (x) = x^{2} - 2 x - 2

f (x) = e^{x} - e^{- x}

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

y^{^{'}} = \frac{x ^{3} y ^{3} + 2 x ^{3}}{y ^{2}}

x \to 7 + lim (\frac{6 x}{x - 7})

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}