How do you differentiate #y=sqrt[(x^2-1)/(x^2+1)]#?

1 Answer
George C.
Mar 12, 2017

#(dy)/(dx) = (2x)/((x^2+1)^2sqrt((x^2-1)/(x^2+1)))#

Explanation:

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

#color(white)(y) = sqrt((x^2+1-2)/(x^2+1))#

#color(white)(y) = sqrt(1-2/(x^2+1))#

#color(white)(y) = (1-2/(x^2+1))^(1/2)#

So using the power rule and chain rule twice we find:

#(dy)/(dx) = 1/2(1-2/(x^2+1))^(-1/2)*d/(dx)(1-2/(x^2+1))#

#color(white)((dy)/(dx)) = 1/2(1-2/(x^2+1))^(-1/2)*d/(dx)(1-2(x^2+1)^(-1))#

#color(white)((dy)/(dx)) = 1/2(1-2/(x^2+1))^(-1/2)*(0+2(x^2+1)^(-2)(2x))#

#color(white)((dy)/(dx)) = 1/2(1-2/(x^2+1))^(-1/2)*(4x)/(x^2+1)^2#

#color(white)((dy)/(dx)) = (2x)/((x^2+1)^2sqrt(1-2/(x^2+1)))#

#color(white)((dy)/(dx)) = (2x)/((x^2+1)^2sqrt((x^2-1)/(x^2+1)))#

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