notation
Does math have an incorrect notation / syntax? I don't mean writing misaligned notation (google), but when you take something like a number to powers to powers to powers, $${{2^2}^2}^3$$ (I was told this is incorrect notation by a teacher). Is it really incorrect, or does it just need to be simplified with parentheses? Do people write maths like this?
a radical expression with the root being a radical expression? $$\sqrt[\sqrt{2^3}]{2}$$
$f'$ is a function, so $f'(2x + 1)$ denotes $f'$ applied to $2x + 1$, or $\frac{df}{dx}(2x + 1)$. For example if $f = x^2$ then $f' = 2x$ and $f'(2x + 1) = 4x + 2$.
$(f(x^2 + 1))'$ is the derivative of the function $f(x^2 + 1)$, which is $2x f'(x^2 + 1)$ by the chain rule.
This question highlights a weakness of the $'$ notation, which is that it always comes with an implied variable with respect to which you're differentiating. If this variable is clear from context there's no problem, but sometimes it isn't.
It is not "fundamentally incorrect." It is informal and intuitive. It is true that Leibniz notation does not rigorously prove anything, but often it is more important (in my opinion) to have an intuitive understanding of why things make sense than of how to prove them formally.
A derivative is a measure of how much one variable changes in response to a change in another variable. So the symbol "$dy$" is the change in y which - taking y as a dependent variable - has been caused by a change in, say, $x$. That change we will call $\Delta x$ or $dx$. The ratio of the two changes $$\frac{dy}{dx}$$ represents how much y is altered in proportion to alterations in x (for small changes).
In this light, the chain rule should be obvious. Say z depends on y, and y in turn depends on x. $$z=z(y),\hspace{2mm} y=y(x)$$ If we change x, what happends to z? Well, when y changes by $\Delta y$, z will change by $$\frac{dz}{dy}\Delta y$$ So how much has y changed? It has changed by $$\Delta y=\frac{dy}{dx}\Delta x$$ In total then, the change in z is $$\Delta z=\frac{dz}{dy}\frac{dy}{dx}\Delta x$$ The ratio of these changes is just $$\frac{\Delta z}{\Delta x}=\frac{dz}{dy}\frac{dy}{dx}$$ In other words, we find the rate of change of z by using its derivative with respect to y, and then incorporating how much y has actually been effected by the independent variable x.
We could even make things more fancy. What would $$\frac{d\big(\sin(x)\big)}{d\big(\sin(x)\big)}$$ equal (assuming that you accept it as coherent)? Clearly if $\sin(x)$ changes by some amount, the dependent function $\sin(x)$ will change by an equal amount. So the derivative is $1$. It is not a mistake that $$\frac{dx}{dx}=1$$ The changes are equivalent and so "cancel." You can also interpret this as saying the "line $y=x$ has a slope of $1$ at all points," but the Leibniz notation is designed for another interpretation.
Inverse functions pose a similar issue. You probably know that $$\frac{d}{dx}\left(f^{-1}(x)\right)=\frac{1}{f'\left(f^{-1}(x)\right)}$$ With differentials, it is easier to see why this makes sense. Clearly $$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$ (Remember, I don't say "clearly" because that is simple algebra - I say it because if you understand what the derivative really means, you should see why an application of said algebra is logically valid.) The change in x and the change in y, as long as the function is actually invertible, remain the same. The inverse function will have a derivative that is the reciprocal of the original function.
-----Are there "counterexamples" to this version of differentials?-----
I would argue that there are not (and bring them on if you think you've found one!). The one which is most commonly exhibited is the following interesting formula. Let $$z=f(x,y)=0$$ Then $$\frac{dy}{dx}=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$ If we were to "cancel" the differentials, we would mistakenly deduce the false claim $$\frac{dy}{dx}=-\frac{dy}{dx}$$ where I have changed partials back to normal derivatives.
This example, however, is based on a simple misunderstanding of what the symbols represent. The "$\partial f$" which occurs in $$\frac{\partial f}{\partial x} $$ is the change in $f$ resulting from a change in the first variable $x$. The "$\partial f$" which occurs in $$\frac{\partial f}{\partial y} $$ is the change in $f$ resulting from a completely different change in the second variable $y$. In other words, the two "$\partial f$"s are different quantities. They cannot be canceled. I would add that anti-intuitionists make Leibniz roll over in his grave every time they think he would have fallen for such an elementary error on the basis of what I consider to be the most brilliant notation ever devised.
Moreover, a simple 'proof' of the equation follows from basic reasoning about differentials. We have that $$\Delta f=\frac{\partial f}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y$$ Using the fact that $$\Delta y=\frac{dy}{dx}\Delta x$$ we get $$\Delta f=\Delta x\left(\frac{\partial f}{\partial x}+\frac{dy}{dx}\frac{\partial f}{\partial y}\right)$$ Since $f(x,y)=0$ over the curve, it will not change. Hence $\Delta f=0$ and the equation follows immediately.
A derivative divides two quantities which are both very small. An integral does just the opposite: it multiplies two quantities, one of which is large (the sum), and one of which is small (the "$\Delta x$").
The integral $$\int$$ looks like an elongated "S" on purpose - it is meant to be a "sum." It is a sum of quantities multiplied by the intervals over which they are evaluated: $$\sum{f(x_i)\Delta x_i}$$ Note that the $\Delta x$ still means the same thing; it is how much we have allowed $x$ to change. So what would happen if we integrated (summed) differentials (differences)? In other words, what is $$\int{d\big(f(x)\big)}$$ Well, if we add up all the changes over the sub-intervals, we get the net (or "total") change. Algebraically, $$\sum{\Delta f(x_i)}=\big(f(x_1)-f(x_0)\big)+\big(f(x_2)-f(x_1)\big)+\cdots +\big(f(x_n)-f(x_{n-1})\big)=f(x_n)-f(x_0)$$ because the sum telescopes. In other words, we have that $$\int{d\big(f(x)\big)}=f(x)$$ and if bounds are specified, we take the change in $f$ at the bounds (that is, $f(b)-f(a)$). The more familiar form of this theorem is $$\int{f'(x)\hspace{1mm}dx}=f(b)-f(a)$$ Once again, $$d\big(f(x)\big)=f'(x)dx$$ It all clicks!
Now let me re-write your example to see if it makes more sense. We can say that $$\int{\cos(3x)dx}=\frac{1}{3}\int{\cos(3x)3dx}=\frac{1}{3}\int{\cos(3x)d(3x)}=\frac{1}{3}\sin(3x)$$ Or another "fancy" one: We could say that $$\int{2\sin(x)d\big(\sin(x)\big)}=\sin^2(x)$$ Why? Because we are adding up changes in $\sin^2(x)$ - observe that $d(\sin^2(x))=2\sin(x)d(\sin(x))$.
If you have had multi-variable calculus, we can draw the examples further. Suppose we are asked to compute a line integral of the form $$\int{xdy+ydx}$$ The "Leibniz" way to solve this problem is as follows (think about the product rule). $$\int{xdy+ydx}=\int{d(xy)}=xy$$ The "normal" way to solve it would go something like this: We check that $$N_x-M_y=0-0=0$$ so the field is conservative. We integrate $f_x=y$ to get $$f(x,y)=xy+\phi(y)$$ Then $$f_y=x+\phi'(y)=x$$ so (after some work) for $f(x,y)=xy+c,\hspace{3mm} c\in\mathbb{R}$, we have $\nabla f=(y,x)=(M,N)$ and therefore the integral is $f=xy+c$.
Or take the famous Integration By Parts formula $$\int{udv}=uv-\int{vdu}$$ Most people don't seem to have any intuition for why this is true. However, if we manipulate it slightly we get $$\int{udv}+\int{vdu}=uv$$ Does that not look suspiciously like what we had just above? Can there really be a connection between integration by parts and vector line integrals?? YES!! They are all based on the same intuitive ideas.
Don't get me wrong: it is immensely important to understand the real definition of limits, and to witness how they can be used to systematically build up calculus into a formal and rigorous structure. But this should not be done at the cost of losing an informal grasp of what calculus is really about, and how it was discovered in the first place (hint: Newton did not have an epiphany involving epsilons and deltas when he saw an apple fall). I do not have space to talk about more objects in calculus, but I believe that, with enough thought, they all turn out to be deeply related and surprisingly intuitive.
Best Answer
Your teacher is mistaken. There is a well-established and universal convention about the meaning of an expression like $$2^{2^{2^3}}$$it is always understood to mean $$2^{\left(2^\left(2^3\right)\right)} =2^{2^8} = 2^{256}$$ People can and do write expressions like these. For example this paper, "Analog of the Skewes Number for Twin Primes", by Marek Wolf, contains the expressions $$10^{10^{10^{10^3}}}\qquad\text{and}\qquad 10^{10^{529.7}}$$on the first page, with no further explanation. Similarly "Some Rapidly Growing Functions" by Craig Smoryński has $$10^{10^{10^{34}}} < e^{e^{e^{e^{4.369}}}}$$ and similar expressions. (I picked these two papers arbitrarily; they were the first two hits in Google Scholar for "Skewes' Number".)
There is a good reason for the convention about what $a^{b^c}$ means: $a^{b^c}$ could be understood as either $a^\left({b^c}\right)$ or as $\left(a^b\right)^c$. But if it were understood as $\left(a^b\right)^c$, one would never need to write $a^{b^c}$, since it would be equal to $a^{bc}$. So it is always understood as $a^\left({b^c}\right)$.
Nobody ever writes $$\sqrt[\sqrt{2^3}]2$$ even though its meaning is clear. Partly this is because it would have been difficult to typeset with old-fashioned metal type, so there is a tradition of expressing this differently. And partly it is because it looks bad.
Since by definition, $$\sqrt[a]b = b^{1/a},$$ one would almost always write something like $$(2^{1/2})^{1/2^{3/2}}$$ instead, at which point it would become clear that the expression could be simplified to $$2^{(1/2)(1/2^{3/2})} = 2^{1/2^{5/2}} = 2^{2^{-5/2}}.$$ Good notation enables and encourages this sort of simplification; bad notation obscures and impedes it.