Find y' for y = x arcsin x
Find $y'$ for $y = x\arcsin x$.
Explanation of Method
A product of an algebraic and an inverse trigonometric function.
Step 1: Identify the factors and apply the product rule
For $y = uv$ recall $y' = u'v + uv'$. Here
\[ u = x,\quad u' = 1, \qquad v = \arcsin x,\quad v' = \frac{1}{\sqrt{1-x^{2}}}. \]
Step 2: Substitute
\[ y' = 1\cdot\arcsin x + x\cdot\frac{1}{\sqrt{1-x^{2}}} = \arcsin x + \frac{x}{\sqrt{1-x^{2}}}. \]
Step 3: Verification
\[x = 0:\quad y'(0) = \arcsin 0 + 0 = 0, \text{ matching the minimum } y(0) = 0 \;\;\checkmark\]
\[x = \tfrac12:\quad y' = \frac{\pi}{6} + \frac{1/2}{\sqrt{3}/2} = \frac{\pi}{6} + \frac{1}{\sqrt3} \approx 1.1009\]
\[\frac{y(0.5+10^{-5})-y(0.5-10^{-5})}{2\times 10^{-5}} \approx 1.1009 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \arcsin x + \dfrac{x}{\sqrt{1-x^{2}}}, \qquad |x| < 1\,}\]
Remark
Remark. Both terms of $y'$ are odd functions, so $y'$ is odd and $y = x\arcsin x$ is even — visible in the graph's symmetry about the $y$-axis. Since $y \geq 0$ with equality only at $x = 0$, the vanishing derivative there marks the minimum.