## Problem Statement

Find the derivative.

$$
y=\left(a^{2}+b^{2}\right)^{-\frac{1}{2}} \cdot \arcsin \left(\frac{\sqrt{a^{2}+b^{2}} \cdot \sin x}{b}\right)
$$
## Solution

$$
\begin{aligned}
& y^{\prime}=\left(\left(a^{2}+b^{2}\right)^{-\frac{1}{2}} \cdot \arcsin \left(\frac{\sqrt{a^{2}+b^{2}} \cdot \sin x}{b}\right)\right)^{\prime}= \\
& =\left(a^{2}+b^{2}\right)^{-\frac{1}{2}} \cdot \frac{1}{\sqrt{1-\left(\frac{\sqrt{a^{2}+b^{2} \cdot \sin x}}{b}\right)^{2}}} \cdot\left(\frac{\sqrt{a^{2}+b^{2}} \cdot \sin x}{b}\right)^{\prime}= \\
& =\frac{1}{\sqrt{a^{2}+b^{2}}} \cdot \frac{b}{\sqrt{b^{2}-\left(a^{2}+b^{2}\right) \cdot \sin ^{2} x}} \cdot \frac{\sqrt{a^{2}+b^{2}}}{b} \cdot \cos x= \\
& =\frac{\cos x}{\sqrt{b^{2} \cdot\left(1-\sin ^{2} x\right)-a^{2} \cdot \sin ^{2} x}}=\frac{\cos x}{\sqrt{b^{2} \cdot \cos ^{2} x-a^{2} \cdot \sin ^{2} x}}
\end{aligned}
$$

## Problem Kuznetsov Differentiation 15-6