Find y' for y = ln (sqrt(1+e^x)-1) - ln (sqrt(1+e^x)+1)

Find $y'$ for $y = \ln\!\left(\sqrt{1+e^{x}}-1\right) - \ln\!\left(\sqrt{1+e^{x}}+1\right)$.

Step 1: Introduce a shorthand

Let $s = \sqrt{1+e^{x}}$, so that

\[ s' = \frac{e^{x}}{2\sqrt{1+e^{x}}} = \frac{e^{x}}{2s}, \qquad\text{and}\qquad y = \ln(s-1) - \ln(s+1). \]

Step 2: Differentiate

Each logarithm contributes $s'$ divided by its own argument: \[y' = \frac{s'}{s-1} - \frac{s'}{s+1} = s'\!\left(\frac{1}{s-1}-\frac{1}{s+1}\right) = s'\cdot\frac{(s+1)-(s-1)}{s^{2}-1} = \frac{2s'}{s^{2}-1}.\]

Step 3: Substitute back

The key simplification is that $s^{2}-1 = \bigl(1+e^{x}\bigr)-1 = e^{x}$, which cancels the $e^{x}$ in $s'$: \[y' = \frac{2}{e^{x}}\cdot\frac{e^{x}}{2s} = \frac{1}{s} = \frac{1}{\sqrt{1+e^{x}}}.\]

Step 4: Verification

\[x = 0:\quad s = \sqrt2 \approx 1.41421 \;\Longrightarrow\; y' = \frac{1}{\sqrt2} \approx 0.70711\]

\[\frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 0.70711 \;\;\checkmark\]

Step 5: Domain

Since $e^{x} \gt 0$ we have $s \gt 1$, so both logarithms are defined for every real $x$ — the argument $s-1$ is positive throughout.