# What is the derivative of #y = arctan(x - sqrt(1+x^2))#?

Thus:

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To find the derivative of ( y = \arctan(x - \sqrt{1+x^2}) ), we'll apply the chain rule.

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

( \frac{du}{dx} = 1 - \frac{1}{2\sqrt{1+x^2}} \cdot 2x = 1 - \frac{x}{\sqrt{1+x^2}} ).

Now, let ( y = \arctan(u) ).

( \frac{dy}{du} = \frac{1}{1 + u^2} ).

Using the chain rule, we multiply ( \frac{dy}{du} ) and ( \frac{du}{dx} ):

( \frac{dy}{dx} = \frac{1}{1 + (x - \sqrt{1+x^2})^2} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}}\right) ).

Simplify the expression:

( \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{1}{2 + 2x^2 - 2x\sqrt{1+x^2}} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}}\right) ).

Thus, the derivative of ( y ) with respect to ( x ) is:

( \frac{dy}{dx} = \frac{1 - \frac{x}{\sqrt{1+x^2}}}{2 + 2x^2 - 2x\sqrt{1+x^2}} ).

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