Find y' for y = arctan e^x - ln sqrt((e^2x)/(e^2x+1))

Find $y'$ for $y = \arctan e^{x} - \ln\sqrt{\dfrac{e^{2x}}{e^{2x}+1}}$.

Step 1: Simplify the logarithm

Pull out the square root, split the quotient, and use $\ln e^{2x} = 2x$: \[\ln\sqrt{\frac{e^{2x}}{e^{2x}+1}} = \frac12\Bigl[\ln e^{2x} - \ln\bigl(e^{2x}+1\bigr)\Bigr] = \frac12\Bigl[2x - \ln\bigl(e^{2x}+1\bigr)\Bigr] = x - \frac12\ln\bigl(e^{2x}+1\bigr).\]

Subtracting this, the function becomes a sum of three elementary terms: $y = \arctan e^{x} - x + \frac12\ln\bigl(e^{2x}+1\bigr)$.

Step 2: Differentiate each term

\[ \frac{d}{dx}\arctan e^{x} = \frac{e^{x}}{1+e^{2x}}, \qquad \frac{d}{dx}(-x) = -1, \qquad \frac{d}{dx}\!\left(\frac12\ln\bigl(e^{2x}+1\bigr)\right) = \frac{e^{2x}}{e^{2x}+1}. \]

Step 3: Combine

The first and third terms share the denominator $1+e^{2x}$: \[y' = \frac{e^{x}+e^{2x}}{1+e^{2x}} - 1 = \frac{e^{x}+e^{2x} - \bigl(1+e^{2x}\bigr)}{1+e^{2x}} = \frac{e^{x}-1}{1+e^{2x}}.\]

Step 4: Verification

\[x = 0:\quad y' = \frac{1-1}{2} = 0, \text{ a critical point} \;\;\checkmark\]

\[x = 1:\quad y' = \frac{e-1}{1+e^{2}} \approx \frac{1.71828}{8.38906} \approx 0.20482; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 0.20482 \;\;\checkmark\]