How do you find the derivative of the function: $y = arctan(x - sqrt(1+x^2)$ ?

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Answer 1 · 2017-11-14T13:04:52

$y'=(sqrt(x^2+1)-x)/[sqrt(x^2+1)*(2x^2-2xsqrt(x^2+1)+2)]$

$y=Arctan[x-sqrt(x^2+1)]$

$tany=x-sqrt(x^2+1)$

$y'*(secy)^2=1-(2x)/[2sqrt(x^2+1)]$

$y'*[(tany)^2+1]=1-x/sqrt(x^2+1)$

$y'*([x-sqrt(x^2+1)]^2+1)=(sqrt(x^2+1)-x)/sqrt(x^2+1)$

$y'*(2x^2-2xsqrt(x^2+1)+2)=(sqrt(x^2+1)-x)/sqrt(x^2+1)$

$y'=(sqrt(x^2+1)-x)/[sqrt(x^2+1)*(2x^2-2xsqrt(x^2+1)+2)]$

How do you find the derivative of the function: y = arctan(x - sqrt(1+x^2)y = arctan(x - sqrt(1+x^2)?