Find y' for y = arcsin 2/x

Find $y'$ for $y = \arcsin\dfrac{2}{x}$.

Step 1: Apply the chain rule

With $u = \dfrac{2}{x}$ and $u' = -\dfrac{2}{x^{2}}$: \[y' = \frac{u'}{\sqrt{1-u^{2}}} = \frac{1}{\sqrt{1-\dfrac{4}{x^{2}}}}\cdot\left(-\frac{2}{x^{2}}\right).\]

Step 2: Simplify the radical

Pulling $x^{2}$ out of the square root produces $|x|$, not $x$ — this is the step to be careful with: $\sqrt{1-\frac{4}{x^{2}}} = \sqrt{\frac{x^{2}-4}{x^{2}}} = \frac{\sqrt{x^{2}-4}}{|x|}$.

Step 3: Combine

Using $\dfrac{|x|}{x^{2}} = \dfrac{1}{|x|}$: \[y' = \frac{|x|}{\sqrt{x^{2}-4}}\cdot\left(-\frac{2}{x^{2}}\right) = -\frac{2}{|x|\sqrt{x^{2}-4}}.\]

Step 4: Verification

\[x = 4:\quad y' = -\frac{2}{4\sqrt{12}} = -\frac{1}{2\sqrt{12}} \approx -0.14434 \;\;\checkmark\]

\[x = -4:\quad y' = -\frac{2}{4\sqrt{12}} \approx -0.14434, \text{ the same value — the absolute value matters} \;\;\checkmark\]