The given function is ( Y = \arctan(x - \sqrt{1 + x^2}) ).

Let ( u = x - \sqrt{1 + x^2} ).

Differentiating both sides with respect to x:

[ \frac{du}{dx} = 1 - \frac{1}{2\sqrt{1+x^2}} \cdot 2x ]

[ \frac{du}{dx} = 1 - \frac{x}{\sqrt{1+x^2}} ]

Now,

[ \frac{dY}{dx} = \frac{d(\arctan u)}{du} \cdot \frac{du}{dx} ]

[ \frac{dY}{dx} = \frac{1}{1+u^2} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}}\right) ]

Since ( u = x - \sqrt{1 + x^2} ), we have

[ \frac{dY}{dx} = \frac{1}{1+(x - \sqrt{1+x^2})^2} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}}\right) ]

[ \frac{dY}{dx} = \frac{1}{1+x^2 - 2x\sqrt{1+x^2} + 1+x^2} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}}\right) ]

[ \frac{dY}{dx} = \frac{1}{2+2x^2 - 2x\sqrt{1+x^2}} \cdot \left(\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}} \right) ]

Hence, the derivative of the function ( Y = \arctan(x - \sqrt{1 + x^2}) ) is given by the above expression.