Find y' for y = x arcsin(ln x)

Find $y'$ for $y = x\arcsin(\ln x)$.

Step 1: Differentiate the second factor

With $u = \ln x$, $u' = \dfrac1x$: \[\bigl(\arcsin(\ln x)\bigr)' = \frac{1}{\sqrt{1-\ln^{2}x}}\cdot\frac{1}{x} = \frac{1}{x\sqrt{1-\ln^{2}x}}.\]

Step 2: Apply the product rule

With $u = x$, $u' = 1$, the factor $x$ cancels against the $\dfrac1x$ above: \[y' = 1\cdot\arcsin(\ln x) + x\cdot\frac{1}{x\sqrt{1-\ln^{2}x}} = \arcsin(\ln x) + \frac{1}{\sqrt{1-\ln^{2}x}}.\]

Step 3: Verification

\[x = 1:\quad \ln 1 = 0 \;\Longrightarrow\; y = 0,\quad y' = 0 + 1 = 1\]

\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]

Step 4: Domain

The arcsine requires $|\ln x| \leq 1$, that is $e^{-1} \leq x \leq e$; the derivative excludes the endpoints, where the radical vanishes and the graph has vertical tangents.