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.
 A: The symbol "$\frac d{dx}$" is used to indicate a single derivative (with respect to $x$).
We treat repeated application of this operator symbolically as "powers" of the operator (as if it were ordinary multiplication by an ordinary fraction), writing "$\frac{d^n}{dx^n}$" to indicate $n$ successive applications of "$\frac d{dx}$".
The notation is peculiar but wholly accepted as traditional. In particular, one might wonder why "$dx^n$" rather than "$(dx)^n$" in the "denominator"; but evidently the "$d$" isn't regarded as an independent factor, rather "$dx$" is regarded as an atomic term.
One eventually accepts it and gets used to the notation.
A: 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. 
A: The correct notation for the sixth derivative is
$$\frac{d^6 y}{dx^6}$$
not $\frac{d^6 y}{d^6x}$. This notation is meant to be suggestive of taking the sixth power of the operator $d/dx$; that is,
$$\frac{d^6 y}{dx^6} = \underbrace{\frac{d}{dx} \cdots \frac{d}{dx}}_{6} y$$
Imagine $dx$ being treated as a single term, of which there are $6$.
A: For convenience, first transform
$$-2\cos^2(x)=-1-\cos(2x).$$
Then the sixth derivative is $$2^6\cos(2x),$$ because $\cos(x)''=-\cos(x)$ and because of the scaling of the variable .
At $x=\dfrac\pi4$, $$0.$$
