Find y' for y = 1/(sqrt(b)) arcsin (x sqrt(b/a)), where

Find $y'$ for $y = \dfrac{1}{\sqrt{b}}\arcsin\!\left(x\sqrt{\dfrac{b}{a}}\right)$, where $a, b \gt 0$.

Step 1: Apply the chain rule

With $u = x\sqrt{\dfrac{b}{a}}$, a constant multiple of $x$, the inner derivative is that same constant: \[u' = \sqrt{\frac{b}{a}}, \qquad y' = \frac{1}{\sqrt{b}}\cdot\frac{1}{\sqrt{1-u^{2}}}\cdot\sqrt{\frac{b}{a}}.\]

Step 2: Cancel the parameter $b$

The factor $\sqrt{b}$ in the numerator cancels the one in the denominator: \[\frac{1}{\sqrt{b}}\cdot\sqrt{\frac{b}{a}} = \frac{1}{\sqrt{a}}, \qquad\text{so}\qquad y' = \frac{1}{\sqrt{a}\,\sqrt{1-\dfrac{b}{a}x^{2}}}.\]

Step 3: Pull $\sqrt{a}$ inside the radical

\[ \sqrt{a}\,\sqrt{1-\frac{b}{a}x^{2}} = \sqrt{a\!\left(1-\frac{b}{a}x^{2}\right)} = \sqrt{a-bx^{2}}. \]

Step 4: Verification

\[a = b = 1:\quad y = \arcsin x \;\Longrightarrow\; y' = \frac{1}{\sqrt{1-x^{2}}} \;\;\checkmark\]

\[a = 4,\ b = 1,\ x = 1:\quad y' = \frac{1}{\sqrt{3}} \approx 0.57735; \qquad \frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 0.57735 \;\;\checkmark\]