1. Problem: Find the derivative (уламжлал) of the function \(y = \frac{\sqrt{x+2}}{\arctan(x-2)}\). 2. Formula and rules: To differentiate a quotient \(\frac{u}{v}\), use the quotient rule: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ where \(u = \sqrt{x+2} = (x+2)^{1/2}\) and \(v = \arctan(x-2)\). 3. Differentiate \(u\): $$\frac{du}{dx} = \frac{1}{2}(x+2)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x+2}}$$ 4. Differentiate \(v\): $$\frac{dv}{dx} = \frac{1}{1+(x-2)^2} \cdot 1 = \frac{1}{1+(x-2)^2}$$ 5. Apply quotient rule: $$y' = \frac{\arctan(x-2) \cdot \frac{1}{2\sqrt{x+2}} - \sqrt{x+2} \cdot \frac{1}{1+(x-2)^2}}{(\arctan(x-2))^2}$$ 6. Final answer: $$\boxed{y' = \frac{\arctan(x-2)}{2\sqrt{x+2}(\arctan(x-2))^2} - \frac{\sqrt{x+2}}{(1+(x-2)^2)(\arctan(x-2))^2}} = \frac{\frac{\arctan(x-2)}{2\sqrt{x+2}} - \frac{\sqrt{x+2}}{1+(x-2)^2}}{(\arctan(x-2))^2}$$ This derivative tells us how the function changes at any point \(x\).