How do you find the derivative of #f(x)= (x^2-1)/x#?

Separate the top and bottom into f and g then find f' and g' and then put it in to the quotient rule

To find the derivative of ( f(x) = \frac{x^2 - 1}{x} ), you can use the quotient rule, which states that if ( u(x) ) and ( v(x) ) are differentiable functions, then the derivative of their quotient ( \frac{u(x)}{v(x)} ) is given by:

[ \left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

In this case, ( u(x) = x^2 - 1 ) and ( v(x) = x ). Thus, the derivative of ( f(x) ) can be calculated as follows:

[ \begin{align*} f'(x) &= \frac{(x^2 - 1)'(x) - (x^2 - 1)(x)'}{(x)'^2} \ &= \frac{(2x)(x) - (x^2 - 1)(1)}{x^2} \ &= \frac{2x^2 - (x^2 - 1)}{x^2} \ &= \frac{2x^2 - x^2 + 1}{x^2} \ &= \frac{x^2 + 1}{x^2} \end{align*} ]

So, the derivative of ( f(x) = \frac{x^2 - 1}{x} ) is ( f'(x) = \frac{x^2 + 1}{x^2} ).