“Man muss immer umkehren”
– Carl Gustav Jacob Jacobi1

This is a short explainer about the chain rule and autograd in PyTorch and JAX, from the perspective of a mathematical user. The Joker

Carl Gustav Jacob Jacobi

(1804 – 1851)
“Die Haare immer nach hinten kehren”
Image source: Wikipedia

There are many, many explanations of this on the web. Many are likely better than this one. I’ll still write my own, which in the spirit of this blog is meant to be written, not read. I’ll also don’t focus on the implementation of the system, just on its observable behavior.

Other, perhaps better sources for the same info are:

The JAX docs especially are delightfully mathematical. There are many more sources on the web. I like the more extensive treatment is in Thomas Frerix’s PhD thesis.

Still, let me add my own spin on the issue. One reason is that many other articles write things like dL/dOutputs and dL/dInputs and generallly use a “variable”-based notation that, while entirely reasonable from an implementation standpoint, would make Spivak sad.

## The chain rule

All of “deep learning”, and most of Physics, depends on this theorem discovered by Leibniz ca. 1676.

### One dimensional functions

Given two differentiable functions $$f,g\from\R\to\R$$, the chain rule says

$(g\circ f)'(x) = g'(f(x))f'(x) \where{x\in\R}. \label{eq:cr1}\tag{1}$

Here, $$(g\circ f)(x) = g(f(x))$$ is the composition, i.e., chained evaluation of first $$y = f(x)$$, then $$g(y)$$.

### Multidimensional functions

It’s one of the wonders of analysis that this rule keeps being correct for differentiable multidimensional functions $$f\from\R^n\to\R^m$$, $$g\from\R^m\to\R^k$$ if “differentiable” is defined correctly. Skipping over some technicalities, this requires $$f$$ and $$g$$ to be totally differentiable (also called Fréchet differentiable), which implies that all components $$f_j\from\R^n\to\R$$ are partially differentiable; the derivative $$f'(a)$$ for $$a\in\R^n$$ can then be shown to be equal to the Jacobian matrix

$f'(a) = J_f(a) := \bigl(\partial_k f_j(a)\bigr)_{\substack{j=1,\ldots,m\\ k=1,\ldots,n}} = \begin{pmatrix} \partial_1 f_1(a) & \cdots & \partial_n f_1(a) \\ \vdots && \vdots \\ \partial_1 f_m(a) & \cdots & \partial_n f_m(a) \end{pmatrix} \in\R^{m\times n}.$

The idea is to view $$f(a) = (f_1(a), \ldots, f_m(a))^\top\in\R^m$$ as a column vector and add one column per partial derivative, i.e., dimension of its input $$a$$.

With this definition, the multidimensional chain rule reads like its 1D version $$\eqref{eq:cr1}$$,

$(g\circ f)'(x) = g'(f(x))\cdot f'(x) \where{x\in\R^n}. \label{eq:crN}\tag{2}$

However, in this case this is the matrix multiplication $$\cdot\from\R^{k\times m}\times\R^{m\times n}\to\R^{k\times n}$$ of $$J_g(f(x))$$ and $$J_f(x)$$ and their order is important. Jacobi would have called this nachdifferenzieren.

Via the nature of this rule it iterates, i.e. the derivative of the composition $$h\circ g\circ f$$ is

$(h\circ g\circ f)(x) = h'(g(f(x)))\cdot g'(f(x))\cdot f'(x) = (h'\circ g\circ f)(x)\cdot (g'\circ f)(x)\cdot f'(x),$

and likewise for a composition of $$n$$ functions

$(f_n\circ \cdots \circ f_1)'(x) = (f_n'\circ f_{n-1}\circ\cdots\circ f_1)(x)\cdot (f_{n-1}'\circ f_{n-2}\circ\cdots\circ f_1)(x) \,\cdots\, (f_2'\circ f_1)(x) \cdot f_1'(x). \label{eq:crNn}\tag{3}$

Note that subscripts no longer mean components here, we are talking about $$n$$ functions, each with multidimensional inputs and outputs.

### Backprop

Now, while matrix multiplication isn’t commutative, meaning in general $$AB \ne BA$$, it is associative: For a product of several matrices $$ABC$$, it does not matter if one computes $$(AB)C$$ or $$A(BC)$$; this is what makes the notation $$ABC$$ sensible in the first place. However, this only means the same output is produced by those two alternatives. It does not mean the same amount of “work” (or “compute”) went into either case.

Counting scalar operations, a product $$AB$$ with $$A\in\R^{k\times m}$$ and $$B\in\R^{m\times n}$$ takes $$nkm$$ multiplications and $$nk(m-1)$$ additions, since each entry in the output matrix is the sum of $$m$$ multiplications. In a chain of matmuls like $$ABCDEFG\cdots$$, it’s clear that some groupings like $$((AB)(CD))(E(FG))\cdots$$ might be better than others. In particular, there may well be better and worse ways of computing the Jacobian $$\eqref{eq:crNn}$$. However, to quote Wikipedia:

The problem of computing a full Jacobian of $$f\from\R^n\to\R^m$$ with a minimum number of arithmetic operations is known as the optimal Jacobian accumulation (OJA) problem, which is NP-complete.

The good news is that in important special cases, this problem is easy. In particular, if the first matrix in the product has only one row, or the last only one column, it makes sense to compute the product “from the left” or “from the right”, respectively. And those cases are not particularly pathological either, as they correspond to either the final function $$f_n$$ mapping to a scalar or the whole composition depending only on a scalar input $$x\in\R$$. The former case is especially important as that’s what happens whenever we compute the gradients of a scalar loss function, such as in bascially all cases where neural networks are used.2 The corresponding orders of multiplications in the chain rule $$\eqref{eq:crNn}$$ are known as reverse mode or forward mode, respectively. Ignoring some distinctions not relevant here, reverse mode is also known as backpropagation, or backprop for short.

In this case, the image of $$f_n$$ is one dimensional, and therefore its Jacobian matrix $$f_n'$$ has only one row – it’s a row vector. After multiplication with the next Jacobian $$f_{n-1}'$$, this property is preserved: All matmuls in the chain turn into vector-times-matrix. Specifically, they are vector-times-Jacobian-matrix, more commonly known as a vector-Jacobian product, or VJP. To illustrate:

$\begin{pmatrix} \unicode{x2E3B} \end{pmatrix}_1 \begin{pmatrix} | & | & | \\ | & | & | \\ | & | & | \end{pmatrix}_2 \begin{pmatrix} | & | & | \\ | & | & | \\ | & | & | \end{pmatrix}_3 \cdots \begin{pmatrix} | & | & | \\ | & | & | \\ | & | & | \end{pmatrix}_n = \begin{pmatrix} \unicode{x2E3B} \end{pmatrix} \begin{pmatrix} | & | & | \\ | & | & | \\ | & | & | \end{pmatrix}_3 \cdots \begin{pmatrix} | & | & | \\ | & | & | \\ | & | & | \end{pmatrix}_n = \text{etc.}$

What this means is that for neural network applications, a system like PyTorch or JAX doesn’t need to actually compute full Jacobians – all it needs are vector-Jacobian products. It turns out that (classic) PyTorch does and can in fact do nothing else.

Also notice that the row vector $$\begin{pmatrix}\unicode{x2E3B}\end{pmatrix}$$ multiplied from the left is “output-shaped” from the perspective of the Jacobian it gets multiplied to. This is the reason grad_output in PyTorch always has the shape of the operation’s output and why torch.Tensor.backward receives an argument described as the “gradient w.r.t. the tensor”.

You may wonder what it means for it to have a specific shape vs just being a “flat” row vector as it is here. I certainly wondered about this; the answer is below.

## In PyTorch

Here’s a very simple PyTorch example (lifted from here):

import torch

y = torch.empty(3)
y = x ** 2
y = x ** 2 + 5 * x ** 2
y = 3 * x

v = torch.tensor([1.0, 2.0, 3.0])
y.backward(v)  # VJP.

print("y:", y)

# Manual computation.
dydx = torch.tensor(
[
[2 * x, 0],
[2 * x, 10 * x],
[0, 3],
]
)



In PyTorch. Tensor.backward computes the backward pass via vector-Jacobian products. If the tensor in question is a scalar, an implicit 1 is assumed, otherwise one has to supply a tensor of the same shape which is used as the vector in the VJP, as in the example above.

## But wait, tensor or vector?

For the typical mathematian, the term “tensor” for the multidimensional array in PyTorch and JAX is a bit on an acquired (or not) taste – but to be fair, so is anything about the real tensor product $$\otimes$$ as well.

The situation here seems especially confusing – the chain rule from multidimensional calculus makes a specific point of what’s a row and what’s a column and treats functions $$\R^n\to\R^m$$ by introducing $$m\times n$$ matrices. In PyTorch, functions depend on and produce one (or several) multidimensional “tensors”. What gives?

The answer turns out to be relatively simple: The math part of the backward pass doesn’t depend on these tensor shapes in any deep way. Instead, it does the equivalent of “reshaping” everything into a vector. Example:

x = torch.tensor(
[
[1.0, 2.0],
[3.0, 4.0],
],
)

y = torch.empty((3, 2))
y[0, 0] = x[0, 0] ** 2
y[1, 0] = x[0, 0] ** 2 + 5 * x[0, 1] ** 2
y[2, 0] = 3 * x[1, 0]
y[0, 1] = x[1, 1]
y[1, 1] = 0
y[2, 1] = torch.sum(x)

v = torch.tensor(
[
[1.0, 2.0],
[3.0, 4.0],
[5.0, 6.0],
]
)

print("y:", y)
y.backward(v)  # VJP.


What did this even compute? It’s the equivalent of reshaping inputs and outputs to be vectors, then applying the standard calculus from above:

dydx = torch.tensor(  # y.reshape(-1) x x.reshape(-1) Jacobian matrix.
[
[2 * x[0, 0], 0, 0, 0],  # y[0, 0]
[0, 0, 0, 1],  # y[0, 1]
[2 * x[0, 0], 10 * x[0, 1], 0, 0],  # y[1, 0]
[0, 0, 0, 0],  # y[1, 1]
[0, 0, 3, 0],  # y[2, 0]
[1, 1, 1, 1],  # y[2, 1]
]
)

print("dydx", dydx)
assert torch.equal(
(v.reshape(-1) @ dydx).reshape(x.shape),
)


To be clear: PyTorch does not actually compute the Jacobian only to multiply it from the left with this vector, but what it does has the same output as this less efficient code.

## In JAX

PyTorch is great. But it’s not necessarily principled. This tweet expresses this somewhat more aggressively:

JAX is, arguably, different, at least on the first account. It comes with vector-Jacobian products, Jacobian-vector products, and also the option to compute full Jacobians if required. I couldn’t write up the details better than the JAX Autodiff Cookbook does, so I won’t try. To quote just one relevant portion, adjusted slightly to our notation:

The JAX function vjp can take a Python function for evaluating $$f$$ and give us back a Python function for evaluating the VJP $$(x, v)\mapsto(f(x), v^\top f'(x))$$.

## So what does it really do?

For the autodiff details, read either Zachary DeVito’s colab with an example implementation of what PyTorch does, or the JAX Autodiff Cookbook, or ideally both.

But to round things out, let’s look at how one defines a “custom function” with its own forward and backward pass in PyTorch.

We’ll take elementwise multiplication as our first example; the fancy mathematics name of this simple operation is Hadamard product (also known as Schur product – their motto was “name, always name3). We’ll denote it by $$\odot.$$ To fit it within the standard calculus above, we reinterpret $$\odot\from\R^n\times\R^n\to\R^n$$ as $$\odot\from\R^{2n}\to\R^n$$, multiplying the first and second “half” of its single-vector input, i.e., $$\R^{2n}\ni x\mapsto\odot(x) = (x_jx_{n+j})_{j=1,\ldots,n}\in\R^n.$$ Its derivative is

$\odot'(x) = \begin{pmatrix} x_{n+1} & & & & x_1 & & & \\ & x_{n+2} & & & & x_2 & & \\ & & {\lower 3pt\smash{\ddots}} & & & & {\lower 3pt\smash{\ddots}} & \\ & & & x_{2n} & & & & x_n \end{pmatrix} \in\R^{n\times 2n}, \where{x\in\R^{2n}}$

where empty cells are zeros. It should be immediately obvious that there’s no need to “materialize” these diagonal Jacobians. In fact, given a vector $$v = (v_1, \ldots, v_{2n})\in\R^{2n}$$, the VJP here is just

$v\cdot \odot'(x) = (v_1x_{n+1}, \ldots, v_nx_{2n}, v_{n+1}x_1, \ldots, v_{2n}x_n) \in \R^{2n}.$

(And that already seems like a needless complication of such a simple thing als elementwise multiplication!)

The PyTorch version of this is … well, it’s just x * y, and PyTorch knows full well how to compute and differentiate that. But if we wanted for some reason to add this from scratch, it might look like this:

class HadamardProduct(torch.autograd.Function):
"""Computes lhs * rhs and its backward pass."""
@staticmethod
def forward(ctx, lhs, rhs):
ctx.save_for_backward(lhs, rhs)
return lhs * rhs

@staticmethod
lhs, rhs = ctx.saved_tensors



The ctx argument is used to stash the tensors required for the backward pass somewhere (in Zachary’s colab, this is done via a closure) and the grad_output argument is the output-shaped “vector” of the VJP. It’s a single argument in this case since the function has a single output. PyTorch may reuse this tensor, so it’s important even in cases where the gradient computed has the same shape to “NEVER modify [this argument] in-place”, as the PyTorch docs say.

Testing this:

lhs = torch.tensor([1.0, 2.0], requires_grad=True)

y = mul(lhs, rhs)

print("y =", y)
v = torch.tensor([5.0, 6.0])
y.backward(v)


This example is typical for elementwise operations, where the Jacobian matrices are diagonal and VJPs are just elementwise operations themselves.

To contrast this with another example, let’s look at a real, genuine linear function. Remember that for a matrix $$W\in\R^{m\times n}$$, the function $$\R^n\ni x \mapsto Wx \in \R^m$$ is Fréchet differentiable and its derivative is the constant $$W$$.4 Taking $$x$$ as a constant and $$W=(W_{jk})_{j=1,\ldots,m;\; k=1,\ldots,n}$$ as the “dependent variable”, and “reshaping to vector form” as above, the derivative at $$W$$ is

$\begin{pmatrix} x_1 & \cdots & x_n & & & & & & & \\ & & & x_1 & \cdots & x_n & & & & \\ & & & & & & \ddots & & & \\ & & & & & & & x_1 & \cdots & x_n \end{pmatrix} \in\R^{m\times nm}.$

Forming the VJP with a vector $$v\in\R^m$$ and reshaping back to the shape of $$W$$ results in VJP of $$v\cdot x^\top$$, with entries $$(v_jx_k)_{j,k}$$.

The PyTorch code for that is (cf. the PyTorch docs):

class ActuallyLinear(torch.autograd.Function):
@staticmethod
def forward(ctx, input, weight):
ctx.save_for_backward(input, weight)
output = weight @ input
return output

@staticmethod
input, weight = ctx.saved_tensors
# Checks for extra efficiency -- only compute VJP with vector != zero.


Testing this:

W = torch.tensor(
[
[1.0, 2.0],
[3.0, 4.0],
[5.0, 6.0],
],
)

y = ActuallyLinear.apply(x, W)
print("y =", y)
v = torch.tensor([1.0, 2.0, 3.0])
y.backward(v)

# Save for comparing.

x.grad = None  # Reset to zero.

torch_linear = torch.nn.Linear(W.shape, W.shape, bias=False)

# Set weight without touching gradient tape.
torch_linear.weight.data[...] = W
torch_y = torch_linear(x)
torch_y.backward(v)


I could go on, but this is plenty for now. The last thing worth mentioning is that $$\eqref{eq:crNn}$$ doesn’t quite fit the actual situation of deep neural networks, where one takes the gradient of all weights, and each layer $$f_j$$ depends both on the outputs (“activations”) of the previous layer and its own weight, and the inputs of the first layer $$f_1$$ (and potentially some later layers, e.g. for “targets”) are just “data”, which is treated as a parametrization of the functions. One way to apply the Calculus 102 rule above would be to have earlier layers pass on all weights they don’t need as an identity function (with identity matrix for that part of its Jacobian). This can also be written in other ways, but the margin-bottom here is too small to contain it.
2. To illustrate the extremes this is pushed to for large scale cases: A large language model like GPT3 can have billions or even trillions of weights (corresponding to the dimension of $$x\in\R^n$$) but still computes a scalar (one dimensional) loss.