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

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

Step 1: Simplify the logarithm

As in Problem 130, pull out the root and split the quotient: \[\ln\sqrt{\frac{e^{2x}}{e^{2x}-1}} = \frac12\Bigl[2x - \ln\bigl(e^{2x}-1\bigr)\Bigr] = x - \frac12\ln\bigl(e^{2x}-1\bigr),\]

so that

\[ y = \arctan e^{2x} + x - \frac12\ln\bigl(e^{2x}-1\bigr). \]

Step 2: Differentiate each term

Write $E = e^{2x}$ for brevity, noting $\bigl(e^{2x}\bigr)' = 2E$ and $\bigl(e^{2x}\bigr)^{2} = e^{4x}$: \[\frac{d}{dx}\arctan E = \frac{2E}{1+E^{2}}, \qquad \frac{d}{dx}(x) = 1, \qquad \frac{d}{dx}\!\left(-\frac12\ln(E-1)\right) = -\frac{E}{E-1}.\]

Step 3: Combine over a common denominator

The common denominator is $\bigl(1+E^{2}\bigr)(E-1)$: \[y' = \frac{2E(E-1) + \bigl(1+E^{2}\bigr)(E-1) - E\bigl(1+E^{2}\bigr)}{\bigl(1+E^{2}\bigr)(E-1)}.\]

Expanding the numerator term by term: \[\begin{aligned} 2E(E-1) &= 2E^{2}-2E, \\ \bigl(1+E^{2}\bigr)(E-1) &= E-1+E^{3}-E^{2}, \\ -E\bigl(1+E^{2}\bigr) &= -E-E^{3}. \end{aligned}\]

The cubes cancel, and collecting the rest gives

\[ \bigl(2E^{2}-E^{2}\bigr) + \bigl(-2E+E-E\bigr) - 1 = E^{2}-2E-1. \]

Step 4: Verification

\[x = 1:\quad E = e^{2} \approx 7.38906,\ E^{2} \approx 54.59815\]

\[\text{Term by term: } \frac{2E}{1+E^{2}} + 1 - \frac{E}{E-1} \approx 0.26580 + 1 - 1.15653 = 0.10928\]

\[\text{Formula: } \frac{54.59815-14.77811-1}{55.59815\times 6.38906} \approx \frac{38.82004}{355.21970} \approx 0.10928 \;\;\checkmark\]

Step 5: Why no further simplification

The numerator $E^{2}-2E-1 = (E-1)^{2}-2$ is not a perfect square and shares no factor with $E-1$: substituting $E = 1$ gives $-2 \neq 0$. So the fraction is already in lowest terms, and no tidy answer of the form $\dfrac{e^{2x}-1}{1+e^{4x}}$ exists.