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

Question

5645 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-20T21:58:13

$y'=1/(2x^2+2)$

$y=arctan(x-sqrt(1+x^2))$

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

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

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

$cos^2y=1/(1+tan^2y)=1/(1+(x-sqrt(1+x^2))^2$

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

With a bit of tidying up, this becomes:

$y'=1/(2x^2+2)$

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