Find y' for y = ln fourth root of ((x²+x+1)/(x²-x+1)) +

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

Step 1: Expand the logarithm

\[ y = \frac14\ln\bigl(x^{2}+x+1\bigr) - \frac14\ln\bigl(x^{2}-x+1\bigr) + \frac{1}{2\sqrt3}\!\left(\arctan\frac{2x+1}{\sqrt3} + \arctan\frac{2x-1}{\sqrt3}\right). \]

Step 2: Differentiate the logarithmic terms

\[ \frac{d}{dx}\!\left(\frac14\ln\bigl(x^{2}+x+1\bigr)\right) = \frac{2x+1}{4\bigl(x^{2}+x+1\bigr)}, \qquad \frac{d}{dx}\!\left(-\frac14\ln\bigl(x^{2}-x+1\bigr)\right) = -\frac{2x-1}{4\bigl(x^{2}-x+1\bigr)}. \]

Step 3: Differentiate the arctangents

The key is that the denominators complete the square back into the same quadratics: \[3+(2x+1)^{2} = 4x^{2}+4x+4 = 4\bigl(x^{2}+x+1\bigr), \qquad 3+(2x-1)^{2} = 4\bigl(x^{2}-x+1\bigr).\]

Hence \[\frac{d}{dx}\arctan\frac{2x+1}{\sqrt3} = \frac{2\sqrt3}{3+(2x+1)^{2}} = \frac{\sqrt3}{2\bigl(x^{2}+x+1\bigr)},\]

and similarly for the second. Multiplying by the prefactor $\dfrac{1}{2\sqrt3}$: $\frac{1}{4\bigl(x^{2}+x+1\bigr)} + \frac{1}{4\bigl(x^{2}-x+1\bigr)}$.

Step 4: Collect over each quadratic

\[ y' = \frac{(2x+1)+1}{4\bigl(x^{2}+x+1\bigr)} + \frac{-(2x-1)+1}{4\bigl(x^{2}-x+1\bigr)} = \frac{x+1}{2\bigl(x^{2}+x+1\bigr)} + \frac{1-x}{2\bigl(x^{2}-x+1\bigr)}. \]

Step 5: Combine into one fraction

The product of the two quadratics is $\bigl(x^{2}+x+1\bigr)\bigl(x^{2}-x+1\bigr) = x^{4}+x^{2}+1$, and the numerator telescopes: $(x+1)\bigl(x^{2}-x+1\bigr) = x^{3}+1, \qquad (1-x)\bigl(x^{2}+x+1\bigr) = 1-x^{3}$.

\[ y' = \frac{\bigl(x^{3}+1\bigr) + \bigl(1-x^{3}\bigr)}{2\bigl(x^{4}+x^{2}+1\bigr)} = \frac{2}{2\bigl(x^{4}+x^{2}+1\bigr)} = \frac{1}{x^{4}+x^{2}+1}. \]

Step 6: Verification

\[x = 0:\quad y' = 1; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]

\[x = 1:\quad y' = \frac{1}{3} \approx 0.33333; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 0.33333 \;\;\checkmark\]