what is the difference between $D(g\circ f)(x)$ and $Dg(f(x))$? Let $A$ be an open in $\mathbb R^m$. Let $B$ be open in $\mathbb R^n$. Let $f: A \to  \mathbb R^n$ and $g: B \to  \mathbb R^p$ where $B = f(A)$. If $f$ is differentiable at $a$ and $g$ is differentiable at $f(a) = b$. Then $D(g\circ f)(a) = Dg(f(a) Df(a)$.
I got this theorem in Mukresh's Analysis of Manifolds . I can not understand what is the difference  between $D(g\circ f)(x)$ and $Dg(f(x))$ .
Can anyone please make me understand ?
Thank You in Advance.

I think you are missing parentheses, which makes your question hard to parse. Here is how I parse it:

I can not understand what is the difference between $D(g\circ f)(x)$ and $Dg(f(x))$.

Think of it this way:
$D$ is an operator, which "takes" as input a function and outputs another function. Therefore, if $f$ is the input, then $D(f)$ (often abbreviated $Df$, which does not help with the confusion) is another function, the "output" of $D$ applied to $f$.
Therefore, $D(g\circ f)(x)$ is what you get when applying the function $D(g\circ f)$ on the point $x$. On the other hand, $Dg(f(x))$ is what you get when applying the function $Dg$ (that is, $D(g)$) on the point $f(x)$.
Put differently: $h_1= D(g\circ f)$ and $h_2= D(g)$ are two functions, while $x$ and $y=f(x)$ are two points. With this notation, you are asking about the difference between $h_1(x)$ and $h_2(y)$.
