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

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Answer 1 · 2016-01-26T09:03:27

$2/(x^2-1)$

Use the quotient rule, which states that if $f(x) = g(x)/(h(x))$ then
$f'(x) = (h(x)g'(x) - h'(x)g(x))/(g^2(x))$

$g(x) = (x^2-1)$ so $g'(x) = 2x$
$h(x) = (x+1)$ so $h'(x) = 1$

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

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