Find y' for y = x + 1/2 (x²-1) artanh x, |x| < 1
Find $y'$ for $y = x + \dfrac{1}{2}\bigl(x^{2}-1\bigr)\operatorname{artanh} x$, $|x| \lt 1$.
Explanation of Method
A product whose rational part collapses to a constant.
Step 1: Differentiate the linear term
\[ \frac{d}{dx}(x) = 1. \]
Step 2: Differentiate the product
With $u = x^{2}-1$, $u' = 2x$, $v = \operatorname{artanh} x$, $v' = \dfrac{1}{1-x^{2}}$: \[\frac{d}{dx}\!\left[\frac12\bigl(x^{2}-1\bigr)\operatorname{artanh} x\right] = \frac12\!\left(2x\operatorname{artanh} x + \frac{x^{2}-1}{1-x^{2}}\right).\]
The quotient is exactly $-1$, since $x^{2}-1 = -\bigl(1-x^{2}\bigr)$: $= x\operatorname{artanh} x - \frac12$.
Step 3: Combine
\[ y' = 1 + x\operatorname{artanh} x - \frac12 = x\operatorname{artanh} x + \frac12. \]
Step 4: Verification
\[x = 0:\quad y = 0,\quad y' = 0 + \tfrac12 = \tfrac12; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 0.5 \;\;\checkmark\]
\[x = \tfrac12:\quad \operatorname{artanh}\tfrac12 \approx 0.54931 \;\Longrightarrow\; y' \approx 0.27466 + 0.5 = 0.77466 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = x\operatorname{artanh} x + \dfrac{1}{2}, \qquad |x| < 1\,}\]
Remark
Remark. The factor $x^{2}-1$ was chosen precisely so that it cancels against the denominator $1-x^{2}$ of $\bigl(\operatorname{artanh} x\bigr)'$, leaving a constant. Since $x\operatorname{artanh} x \geq 0$ on $(-1,1)$, the derivative is at least $\tfrac12$ and the function is strictly increasing.