What does $\frac{d^6y}{ dx^6}$ mean? The whole question is that 
If $f(x) = -2cos^2x$, then what is $d^6y \over dx^6$ for x = $\pi/4$?
The key here is what does $d^6y \over dx^6$ mean?
I know that $d^6y \over d^6x$ means 6th derivative of y with respect to x, but I've never seen it before.

If you recall the definition of the derivative, then you can write
\begin{align}
y'(x) &= \lim_{h \to 0} \frac {y(x+h) - y(h)}h \\
y''(x) &= \lim_{h \to 0} \frac {y'(x+h) - y'(x)}h = \lim_{h \to 0} \frac{\frac {y(x+2h) - y(x+h)}h - \frac {y(x+h) - y(x)}h}h = \lim_{h \to 0} \frac {y(x+2h) - 2y(x+h) + y(x)}{h^2}
\end{align}
so you can see, that denominator is in fact raised to the power, and numerator isn't. So, by convention
$$
y'' = \frac {d^2y}{dx^2}
$$
where $d^2y$ means a certain infinitesimal difference operator, and $dx^2$ means actual power.
Same idea holds for higher derivatives. 
