No need for chain rule with $f'(g(x))?$ Let's say I take a simple example of a function, $f(x) = \sin(x)$. I want to calculate $f'(x^2)$. The answer is $\cos(x^2)$. 
However, if $f'(x) = \frac{d}{dx}f(x)$, then shouldn't $f'(x^2) = \frac{d}{dx}f(x^2)$ which is $\frac{d}{dx}\sin(x^2)$ which means $f'(x)$ is $2x\cos(x^2)$?
I'm trying to understand where I went wrong in my understanding, as it seems like you don't actually need to apply chain rule. 
 A: We need to distinguish between the symbol $f'(u)$, which means the function $f'$ at point $u$, that means the value of the derivative of $f$ at point $u$, with the derivative of the function $x\mapsto \sin(x^2)$ at a point. 
Put $h(x):=\sin(x^2)$. Then, by the chain rule,
$h'(x)=2x\cos(x^2)$
We must distinguish between $h'(x)$ and $f'(x^2)$. The first is the derivative of the composition at a point $x$, while the second is the derivative of $f$ at the value $x^2$.
A: That's a matter of symbols and notations.
Usually, the prime is used when you don't risk ambiguity, and this really looks the case where the notation can be unfortunate.
The first derivative that you computed, and called $f'(x^2)$, in standard notation is
$$
\frac{d}{d(x^2)}f(x^2)=\frac{d}{dx^2}\sin(x^2)=\cos(x^2)
$$
If you say $f'(x^2)=\cos(x^2)$, you are taking the prime as a derivative with respect to the variable that is in the parentheses following it.
The second derivative is
$$
\frac{d}{dx}f(x^2)=\frac{dx^2}{dx}\frac{d}{d(x^2)}f(x^2)=2x\cos(x^2)
$$
and that's different. Here derivation is done with respect to $x$, so the chain rule is necessary. Note, the two things are different.
To summarize, the prime notation (and the dot notation, $\dot f$, popular in physics) are to be used only when no ambiguity is possible in identifying the derivation variable. It is a shorthand notation, that in some cases can lead to confusion. Your case is one of such cases, and I'd drop the notation and write the derivative in full form.
A: You have to be extremely careful about the order in which you apply the two operations:


*

*taking the derivative;

*plugging a value into a function.


Unfortunately mathematics has not yet converged to using a universally good notation for differentiation (!) and confusion is understandable (and it gets much much worse once you start doing calculus in multiple dimensions!) Let's look at your example closely:
$f'(x^2)$ usually means "differentiate first, then plug in $x^2$": notice that $x$ is overloaded here, since it is the name of the argument of $f$, and also the name of an (unknown, independent) variable that you are plugging in after differentiating $f$. Let us remove the overload by writing $f(t) = \sin(t)$. Then $f'(x^2)$ really means
$$\left[\frac{d}{dt}f\right](x^2) = [\cos t]_{t= x^2} = \cos x^2.$$
Compare it to $\frac{d}{dx}\left[f(x^2)\right].$ This is asking you to first plug in $x^2$ into $f$, and then take the derivative. This is the case where you need to use the chain rule.
A: It seems that you're confusing the difference between the derivative of the composition $f\circ g$ and the derivative of $f$ evaluated at $g$. In your first example, $f(x)= \sin x$ and $g(x) = x^2$, the derivative of $f$ evaluated at $g$ is 
\begin{equation*}
\left.\frac{d}{dx}\right|_{x^2} f = \cos(x^2). 
\end{equation*}
Here, you've composed the derivative of $f$ with $g$. 
In your second example you're computing the derivative of the composition $f\circ g$. According to the chain rule, 
\begin{equation*}
\left.\frac{d}{dx}\right|_x(f\circ g)
= 
\left.\frac{d}{dx}\right|_{g(x)} f\cdot \left.\frac{d}{dx}\right|_xg
=
\cos(x^2)\cdot 2x. 
\end{equation*}
Many authors use the same notation for these two concepts. You have to use contextual clues to decide what is intended. 
