Find y' for y = ln fourth root of ((x-1)/(x+1)) + 3 ln

Find $y'$ for $y = \ln\sqrt[4]{\frac{x-1}{x+1}} + 3\ln\sqrt[4]{\frac{x^{2}+1}{x^{2}-1}} + \frac{1}{2}\arctan x$.

Step 1: Pull the fourth roots out of the logarithms

\[ y = \frac14\ln\frac{x-1}{x+1} + \frac34\ln\frac{x^{2}+1}{x^{2}-1} + \frac12\arctan x. \]

Step 2: Expand, using $x^{2}-1 = (x-1)(x+1)$

The second logarithm splits into the same two linear factors as the first: \[y = \frac14\ln(x-1) - \frac14\ln(x+1) + \frac34\ln\bigl(x^{2}+1\bigr) - \frac34\ln(x-1) - \frac34\ln(x+1) + \frac12\arctan x.\]

Collecting like terms: $y = -\frac12\ln(x-1) - \ln(x+1) + \frac34\ln\bigl(x^{2}+1\bigr) + \frac12\arctan x$.

Step 3: Differentiate term by term

\[ y' = -\frac{1}{2(x-1)} - \frac{1}{x+1} + \frac34\cdot\frac{2x}{x^{2}+1} + \frac{1}{2\bigl(1+x^{2}\bigr)}. \]

Step 4: Merge the last two terms

They share the denominator $x^{2}+1$: \[\frac{3x}{2\bigl(x^{2}+1\bigr)} + \frac{1}{2\bigl(x^{2}+1\bigr)} = \frac{3x+1}{2\bigl(x^{2}+1\bigr)}.\]

Step 5: Verification

\[x = 2:\quad y' = -\frac{1}{2} - \frac{1}{3} + \frac{7}{10} = -0.5 - 0.33333 + 0.7 = -0.13333\]

\[\frac{y(2+10^{-5})-y(2-10^{-5})}{2\times 10^{-5}} \approx -0.13333 \;\;\checkmark\]