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

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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)#