Consider the function $f(x) = x^2 + 4/x^2$ a) Find$f ^\prime(x)$ b) Find the values of $x$ at which the tangent to the curve is horizontal. So far I have this...
a) $f^\prime(x) = 2x + (0)(x^2)-(4)\dfrac{2x}{(x^2)^2}$ 
$= 2x - \dfrac{8x}{x^4}$
$= \dfrac{2x^5 - 8x}{x^4}$ 
$= \dfrac{2(x^4 - 4)}{x^3}$
I believe I derived this correctly. 
But i am not sure how to do part b). I know the horizontal slope $= 0$ but when i solve it i get $2(x^4 - 4)$ and don't know how to go from there. 
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Your derivative is correct. You could have saved yourself some work by using the power rule.
\begin{align*}
f(x) & = x^2 + \frac{4}{x^2}\\
     & = x^2 + 4x^{-2}
\end{align*}
Using the power rule yields
\begin{align*}
f'(x) & = 2x^1 - 2 \cdot 4x^{-3}\\
      & = 2x - 8x^{-3}\\
      & = 2x - \frac{8}{x^3}
\end{align*}
which is equivalent to your expression
$$f'(x) = \frac{2(x^4 - 4)}{x^3}$$
To find the values of $x$ at which the tangent line is horizontal, set $f'(x) = 0$, which yields
\begin{align*}
\frac{2(x^4 - 4)}{x^3} & = 0\\
2(x^4 - 4) & = 0\\
x^4 - 4 & = 0\\
(x^2)^2 - 2^2 & = 0\\
(x^2 + 2)(x^2 - 2) & = 0\\
\end{align*}
Setting each factor equal to zero yields
\begin{align*}
x^2 + 2 & = 0 & x^2 - 2 & = 0\\
x^2 & = -2 & x^2 & = 2\\
x & = \pm i\sqrt{2} & x & = \pm\sqrt{2}
\end{align*}
Since there can only be a tangent at real values of $x$, we conclude that the only horizontal tangents of the graph occur at $x = \pm\sqrt{2}$.
