Find y' for y = ln arcsin x - (arcsin x)²

Find $y'$ for $y = \ln\arcsin x - (\arcsin x)^{2}$.

Step 1: Introduce a shorthand

Write $A = \arcsin x$, so that $A' = \dfrac{1}{\sqrt{1-x^{2}}}$ and $y = \ln A - A^{2}$.

Step 2: Differentiate each term

\[ \frac{d}{dx}\ln A = \frac{A'}{A} = \frac{1}{A\sqrt{1-x^{2}}}, \qquad \frac{d}{dx}A^{2} = 2A\,A' = \frac{2A}{\sqrt{1-x^{2}}}. \]

Step 3: Combine over a common denominator

Both terms carry the factor $\dfrac{1}{\sqrt{1-x^{2}}}$; putting them over $A\sqrt{1-x^{2}}$: $y' = \frac{1}{A\sqrt{1-x^{2}}} - \frac{2A}{\sqrt{1-x^{2}}} = \frac{1-2A^{2}}{A\sqrt{1-x^{2}}}$.

Step 4: Verification

\[x = \tfrac12:\quad A = \frac{\pi}{6} \approx 0.52360,\ \sqrt{1-x^{2}} \approx 0.86603\]

\[y' \approx \frac{1-2(0.52360)^{2}}{0.52360\times 0.86603} = \frac{0.45169}{0.45345} \approx 0.99612 \;\;\checkmark\]

Step 5: Domain and critical points

The formula needs $|x| \lt 1$ and $\arcsin x \neq 0$, i.e. $x \neq 0$. The derivative vanishes where $2A^{2} = 1$, that is $\arcsin x = \pm\dfrac{1}{\sqrt2}$, giving

\[ x = \pm\sin\frac{1}{\sqrt2} \approx \pm 0.64964. \]