Find y' for y = 1/(arcsin x)
Find $y'$ for $y = \dfrac{1}{\arcsin x}$.
Explanation of Method
Differentiate the reciprocal of an inverse trigonometric function.
Step 1: Identify the parts and apply the quotient rule
For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v - uv'}{v^{2}}$. Here
\[ u = 1,\quad u' = 0, \qquad v = \arcsin x,\quad v' = \frac{1}{\sqrt{1-x^{2}}}. \]
Step 2: Substitute and simplify
The first term of the numerator vanishes because $u' = 0$: \[y' = \frac{0\cdot\arcsin x - 1\cdot\dfrac{1}{\sqrt{1-x^{2}}}}{\bigl(\arcsin x\bigr)^{2}} = -\frac{1}{\sqrt{1-x^{2}}\,\bigl(\arcsin x\bigr)^{2}}.\]
Step 3: Verification
\[\text{Reciprocal rule } \left(\frac{1}{v}\right)' = -\frac{v'}{v^{2}} \text{ gives the same expression in one step} \;\;\checkmark\]
\[x = \tfrac12:\quad \arcsin\tfrac12 = \frac{\pi}{6},\ \sqrt{1-\tfrac14} = \frac{\sqrt3}{2} \;\Longrightarrow\; y' = -\frac{2}{\sqrt3}\cdot\frac{36}{\pi^{2}} \approx -4.212\]
\[\frac{y(0.5+10^{-5})-y(0.5-10^{-5})}{2\times 10^{-5}} \approx -4.212 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\dfrac{1}{\sqrt{1-x^{2}}\,\bigl(\arcsin x\bigr)^{2}}, \qquad |x| < 1,\ x \neq 0\,}\]
Remark
Remark. Two restrictions are at work. The factor $\sqrt{1-x^{2}}$ requires $|x| \lt 1$, and the denominator $\bigl(\arcsin x\bigr)^{2}$ requires $\arcsin x \neq 0$, i.e. $x \neq 0$ — where the original function itself is undefined. The derivative is negative throughout, so $y$ decreases on each of $(-1,0)$ and $(0,1)$.