Find y' for y = arctan x/a - ln sqrt(x^4-a^4), then

Find $y'$ for $y = \arctan\dfrac{x}{a} - \ln\sqrt{x^{4}-a^{4}}$, then evaluate $y'(2a)$. Here $a \gt 0$.

Step 1: Differentiate the arctangent

With $u = \dfrac{x}{a}$, $u' = \dfrac1a$: \[\frac{d}{dx}\arctan\frac{x}{a} = \frac{1}{1+\dfrac{x^{2}}{a^{2}}}\cdot\frac{1}{a} = \frac{a^{2}}{a^{2}+x^{2}}\cdot\frac{1}{a} = \frac{a}{a^{2}+x^{2}}.\]

Step 2: Differentiate the logarithm

Pull the square root out first: $\ln\sqrt{x^{4}-a^{4}} = \tfrac12\ln\bigl(x^{4}-a^{4}\bigr)$.

\[ \frac{d}{dx}\!\left(\frac12\ln\bigl(x^{4}-a^{4}\bigr)\right) = \frac{1}{2}\cdot\frac{4x^{3}}{x^{4}-a^{4}} = \frac{2x^{3}}{x^{4}-a^{4}}. \]

Step 3: Combine

Factoring $x^{4}-a^{4} = \bigl(x^{2}-a^{2}\bigr)\bigl(x^{2}+a^{2}\bigr)$ shows that the two denominators share the factor $x^{2}+a^{2}$: \[y' = \frac{a}{x^{2}+a^{2}} - \frac{2x^{3}}{\bigl(x^{2}-a^{2}\bigr)\bigl(x^{2}+a^{2}\bigr)} = \frac{a\bigl(x^{2}-a^{2}\bigr) - 2x^{3}}{\bigl(x^{2}-a^{2}\bigr)\bigl(x^{2}+a^{2}\bigr)}.\]

Step 4: Evaluate at $x = 2a$

\[ x^{2} = 4a^{2}, \qquad x^{2}-a^{2} = 3a^{2}, \qquad x^{2}+a^{2} = 5a^{2}, \qquad x^{3} = 8a^{3}. \]

\[ y'(2a) = \frac{a\cdot 3a^{2} - 2\cdot 8a^{3}}{3a^{2}\cdot 5a^{2}} = \frac{3a^{3}-16a^{3}}{15a^{4}} = \frac{-13a^{3}}{15a^{4}} = -\frac{13}{15a}. \]

Step 5: Verification

\[a = 1,\ x = 2:\quad y' = \frac{1}{5} - \frac{16}{15} = \frac{3-16}{15} = -\frac{13}{15} \approx -0.86667\]

\[\frac{y(2+10^{-5})-y(2-10^{-5})}{2\times 10^{-5}} \approx -0.86667 \;\;\checkmark\]