what is the difference between $D(g\circ f)(x)$ and $Dg(f(x))$?

Question

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.

Answer 1

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 $h_2(y)$.

Answer 2

We have that

• $D(g\circ f)(x)$ is the Jacobian of $g(f(x))$ at $x$.
• $Dg(f(x))=Dg(y)$ is the Jacobian of $g(y)$ at $y=f(x)$.

To give a concrete example assume

• $f(x,y)=(x^2+y^2,x^2-y^2)$

• $g(u,v)=(u^2,2uv,v^2)$

then

$$Df(x,y)=\begin{bmatrix}\frac{\partial f_1}{\partial x}&\frac{\partial f_1}{\partial y}\\\frac{\partial f_2}{\partial x}&\frac{\partial f_2}{\partial y}\end{bmatrix}=\begin{bmatrix}2x&2y\\2x&-2y\end{bmatrix}$$

$$Dg(u,v)=\begin{bmatrix}\frac{\partial g_1}{\partial u}&\frac{\partial g_1}{\partial v}\\\frac{\partial g_2}{\partial u}&\frac{\partial g_2}{\partial v}\\\frac{\partial g_3}{\partial u}&\frac{\partial g_3}{\partial v}\end{bmatrix}=\begin{bmatrix}2u&0\\2v&2u\\0&2v\end{bmatrix}$$

and then at a specific point $(x_0,y_0)$ and $(u_0,v_0)=(f_1(x_0,y_0),f_2(x_0,y_0)$ we can evaluate the Jacobian for the composition that is

$$D(g\circ f)(x_0,y_0)=Dg(u_0,v_0)\cdot Df(x_0,y_0)=Dg(f_1(x_0,y_0),f_2(x_0,y_0))\cdot Df(x_0,y_0)$$

Answer 3

As I commented, you could rewrite them. Or write $$ \mathrm D(g\circ f)|_{x}, \quad \mathrm Dg|_{f(x)}. $$ Explanations: the differential of $g \circ f$ at $x$, and the differential of $g$ at $f(x)$.