Question

Find the derivative of each function. Then,

determine whether the function has any local
extrema.

Answer 1

Using the quotient rule, the derivative of
`y` is

`y' = ((x^2-1) x' - x(x^2-1)')/((x^2-1)^2) = ((x^2-1) - x(2x))/((x^2-1)^2) = -(x^2+1)/((x^2-1)^2)`

as you're shown in the picture.

If
`y` has any local extrema, they occur at critical points where
`y'=0` or possibly when
`y'` is undefined.

In this case,
`y'` is undefined when
`x=1`, but this is outside the domain of
`y` anyway, so we can ignore that.

That leaves us with the zero-case. For
`x\\eq1`, the denominator is always positive, so the numerator must be zero. But this never happens, since

`-(x^2+1) = 0 \implies x^2 = -1`

but
`x^2\ge0` for all real
`x`. Therefore
`y` has no critical points and thus no extrema.