How do I differentiate $$ (y-X\beta)^T(y - X \beta) $$

with respect to $\beta$. The result I saw was

$$X^T(y - X\beta)$$

  • $\begingroup$ What have you tried so far? Where are you stuck? $\endgroup$
    – Sycorax
    Commented Sep 2, 2020 at 18:06
  • 2
    $\begingroup$ @Tylerr "Solution" in what sense? What you've written is not the derivative of the expression with respect to $\beta$. $\endgroup$
    – Sycorax
    Commented Sep 2, 2020 at 18:25
  • 1
    $\begingroup$ @Tylerr That seems to be the answer to a question that OP didn't ask. $\endgroup$
    – Sycorax
    Commented Sep 2, 2020 at 18:52
  • 1
    $\begingroup$ @EA Lehn here is an ok walkthrough: towardsdatascience.com/… definitely look up matrix calculus rules and algebra rules to follow along. It is actually pretty easy once you grasp the matrix rules, just expand your first equation and take the derivative and solve for beta! Couldn't find one that goes through every single step individually although I am sure one exists. Let me know if you need more help. $\endgroup$
    – Tylerr
    Commented Sep 2, 2020 at 18:53
  • 1
    $\begingroup$ I'm with @Sycorax.. that's a completely different problem to solve. $\endgroup$ Commented Sep 2, 2020 at 19:19

1 Answer 1


Let us assume that you are working in a setup where $y$ is $N \times 1$ and $X$ is $N \times K$ and $\beta$ is $K \times 1$. I prefer to define $e(\beta) := (y - X\beta)$ and similarly the $i$'th component $e_{i}(\beta) = (y - X\beta)_i = y_i - x_i^\top\beta$ where $x_i^\top$ is the $i$'th row of $X$. You should then be able to convince yourself that

$$e(\beta)^\top e(\beta) = \sum_i e_{i}(\beta) e_{i}(\beta),$$

the sum of squared deviations. Now I guess you know how to differentiate with respect to a single variable (read parameter) $\beta_k$ so lets try this

$$\frac{\partial}{\partial \beta_k} e(\beta)^\top e(\beta) = \sum_i\frac{\partial}{\partial \beta_k} [e_{i}(\beta) e_{i}(\beta)],$$

apply the product rule to get

$$= \sum_i \frac{\partial e_i(\beta)}{\partial \beta_k} e_i(\beta) + e_i(\beta) \frac{\partial e_i(\beta)}{\partial \beta_k} = 2 \sum_i \frac{\partial e_i(\beta)}{\partial \beta_k} e_i(\beta),$$

where the final sum here can be written in matrix/vector notation as

$$= 2 \left[\frac{\partial e(\beta)^\top}{\partial \beta_k}\right] e(\beta),$$

all the same derivations can be done differentiating with respect to a column $\beta$ observing the rule that when you differentiate with respect to a column you get a column so

$$\frac{\partial e_i(\beta)}{\partial \beta} = \begin{pmatrix} \frac{\partial e_i(\beta)}{\partial \beta_1}\\ \vdots \\ \frac{\partial e_i(\beta)}{\partial \beta_K} \end{pmatrix}$$

you should then be able to get the rule that

$$\frac{\partial}{\partial \beta} e(\beta)^\top e(\beta) = 2 \left[\frac{\partial e(\beta)^\top}{\partial \beta}\right] e(\beta),$$

so let figure out what $\frac{\partial e(\beta)^\top}{\partial \beta}$ for which we get

$$\frac{\partial e(\beta)^\top}{\partial \beta} = \frac{\partial}{\partial \beta} (e_1(\beta),...,e_N(\beta)) = \left( \frac{\partial e_1(\beta)}{\partial \beta},..., \frac{\partial e_N(\beta)}{\partial \beta}\right)$$ and for each $i$ you have that $\frac{\partial e_{i}(\beta)}{\partial \beta} = -x_i$ so then it is easy to see that $$\frac{\partial e(\beta)^\top}{\partial \beta} = - X^\top$$ and it follows that

$$\frac{\partial}{\partial \beta} e(\beta)^\top e(\beta) = - 2X^\top (y - X\beta).$$

In a context where the writer knows he or she wants to solve $- 2X^\top (y - X\beta) = 0$ he or she may go directly from $$\frac{\partial}{\partial \beta} e(\beta)^\top e(\beta) = 0$$ to $X^\top (y - X\beta) = 0$ leading you to think that the author is implicitly claiming that $$\frac{\partial}{\partial \beta} e(\beta)^\top e(\beta)= X^\top (y - X\beta) $$ which is not the case.


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