Find y' for y = artanh x - arctan x, where |x| < 1
Find $y'$ for $y = \operatorname{artanh} x - \arctan x$, where $|x| \lt 1$.
Explanation of Method
A hyperbolic and a circular inverse function side by side.
Step 1: Recall the standard derivatives
The two formulas differ only in one sign, which is the whole point of the comparison: \[\bigl(\operatorname{artanh} x\bigr)' = \frac{1}{1-x^{2}}\ \ (|x| \lt 1), \qquad \bigl(\arctan x\bigr)' = \frac{1}{1+x^{2}}\ \ (x \in \mathbb{R}).\]
Step 2: Differentiate term by term
\[ y' = \frac{1}{1-x^{2}} - \frac{1}{1+x^{2}}. \]
Step 3: Combine over a common denominator
The product of the denominators is a difference of squares, $\bigl(1-x^{2}\bigr)\bigl(1+x^{2}\bigr) = 1-x^{4}$: \[y' = \frac{\bigl(1+x^{2}\bigr) - \bigl(1-x^{2}\bigr)}{1-x^{4}} = \frac{1+x^{2}-1+x^{2}}{1-x^{4}} = \frac{2x^{2}}{1-x^{4}}.\]
Step 4: Verification
\[x = 0:\quad y'(0) = 0, \text{ and } y(0) = 0 \;\;\checkmark\]
\[x = \tfrac12:\quad y' = \frac{2\cdot\frac14}{1-\frac{1}{16}} = \frac{\frac12}{\frac{15}{16}} = \frac{8}{15} \approx 0.5333\]
\[\frac{y(0.5+10^{-5})-y(0.5-10^{-5})}{2\times 10^{-5}} \approx 0.5333 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{2x^{2}}{1-x^{4}}, \qquad |x| < 1\,}\]
Remark
Remark. On $|x| \lt 1$ both $2x^{2} \geq 0$ and $1-x^{4} \gt 0$, so $y$ is increasing, with a stationary inflection at the origin. As $x \to 1^{-}$ the derivative tends to $+\infty$, matching the vertical asymptote of $\operatorname{artanh} x = \tfrac12\ln\dfrac{1+x}{1-x}$ there, while $\arctan x$ stays perfectly finite. Compare Problem 59, whose derivative $\dfrac{2x^{2}}{\bigl(1+x^{2}\bigr)^{2}}$ has the same numerator.