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

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Answer 1 · 2016-01-15T11:06:47

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

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

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

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

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