Certainly! The chain rule is a fundamental concept in calculus used to differentiate composite functions. It allows us to find the derivative of a function that is composed of multiple functions. Let's break it down and see how it's applied to find the gradient in this context.

Chain Rule Overview

Suppose you have a composite function $f(x) = h(g(x))$. The chain rule states that the derivative of $f(x)$ with respect to $x$ is:

$\frac{d}{dx} f(x) = \frac{d}{dx} h(g(x)) = h'(g(x)) \cdot g'(x)$

This means that you differentiate the outer function $h$ with respect to its argument $g(x)$, and then multiply it by the derivative of the inner function $g(x)$ with respect to $x$.

Application in Gradient Computation

Let's apply this to the gradient computation for the function:

$g(z) = \frac{1}{2} \|Az - b\|_2^2$

Here, the function $g(z)$ is composed of an outer function $h(u) = \frac{1}{2} \|u\|_2^2$, where $u = Az - b$, and the inner function $u(z) = Az - b$. We need to find the gradient of $g(z)$ with respect to $z$.

Step 1: Differentiate the outer function

The outer function is $h(u) = \frac{1}{2} \|u\|_2^2$. The gradient of $h(u)$ with respect to $u$ is:

$\frac{\partial h(u)}{\partial u} = u$

Step 2: Differentiate the inner function

The inner function is $u(z) = Az - b$. To find the gradient of $u(z)$ with respect to $z$:

$\frac{\partial u(z)}{\partial z} = \frac{\partial}{\partial z} (Az - b) = A$

Note that $A$ is a matrix, and $b$ is a constant vector, so their derivatives with respect to $z$ follow linearity rules. Specifically, the derivative of $Az$ with respect to $z$ is $A$, and the derivative of $b$ is zero since $b$ does not depend on $z$.

Step 3: Apply the Chain Rule

Now, apply the chain rule. The gradient of $g(z)$ with respect to $z$ is:

$\nabla_z g(z) = \frac{\partial g(z)}{\partial z} = \frac{\partial h(u)}{\partial u} \cdot \frac{\partial u(z)}{\partial z}$

Substituting the gradients we found:

$\nabla_z g(z) = u \cdot A = (Az - b) \cdot A = A^T(Az - b)$

Summary

The chain rule allowed us to break down the gradient computation into manageable parts:

1. Outer function $h(u) = \frac{1}{2} \|u\|_2^2$: Its gradient with respect to $u$ is $u = Az - b$.
2. Inner function $u(z) = Az - b$: Its gradient with respect to $z$ is $A$.
3. Chain rule application: The gradient of the composite function is $A^T(Az - b)$.

Why the Chain Rule Works

The chain rule works because differentiation is a linear operation that can be distributed over functions that are composed in a nested manner. By breaking down the operation, we ensure that we correctly account for how changes in the inner function propagate through the outer function.

Would you like to explore more on this topic or any specific part of it?

---

Here are some further questions that might be interesting:

1. How is the chain rule applied in higher dimensions for multivariable functions?
2. Can you provide examples of the chain rule in physics or other applied fields?
3. How does the chain rule relate to the concept of the Jacobian matrix in vector calculus?
4. What are common mistakes when applying the chain rule in complex functions?
5. How does the chain rule extend to functions that involve matrices and tensors?

Tip: When applying the chain rule, always carefully identify the inner and outer functions. This step is crucial for correctly differentiating composite functions.