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

1 Answer
Jan 15, 2016

Answer:

#(x(-x^2 +3))/(sqrt(x^2-1)*(x^2+1)^2)#

Explanation:

The derivative can be found by using the quotient rule, which states that if #f(x) = g(x)/(h(x))# then
#f'(x) = (h(x)g'(x) - h'(x)g(x))/(h^2(x))#

In this case #g(x) = (x^2-1)^(1/2)#
and #h(x) = (x^2+1)#

Using the chain rule, #g'(x) = 1/2(x^2 - 1) ^(-1/2)*2x = x/sqrt(x^2-1)#
#h'(x) = 2x#

Then #f'(x) = ((x^2+1)*(x/sqrt(x^2-1)) -2x*sqrt(x^2-1))/(x^2+1)^2#

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

#=(x^3+x-2x^3+2x)/(sqrt(x^2-1)*(x^2+1)^2) = (x(-x^2 +3))/(sqrt(x^2-1)*(x^2+1)^2)#