Find derivative of arctan (x - √(1 + x²) )

I got x / [ √(1 + x²) - x√(1 + x²) - 1 ] for the answer.

Is it correct, if not can you show me how to do it step by step?

Let,
y = arctan (x - √(1 + x²) ),
SO,
tan y = (x - √(1 + x²) ),

sec^2y*dy/dx = 1 - x / √(1+x^2),

dy/dx = [1/sec^2y]*[√(1+x^2) -x] / √(1+x^2),
dy/dx = [1/(1+tan^2y]*[√(1+x^2) -x] / √(1+x^2),
dy/dx = [√(1+x^2) - x] / √(1+x^2)*[1 + (x-√(1+x^2))^2],
OR,
dy/dx = x / [ √(1 + x²) - x√(1 + x²) - 1 ] >====================< ANSWER
Your Answer is CORRECT .......................