This is a quick question about syntax.

I have the function:

$$\hat{x_i} = \frac{x_i - \mu_\beta}{\sqrt{\sigma^2_\beta+\epsilon}}$$

where $\sigma^2_\beta$ is an intermediate function:

$$\sigma^2_\beta = \frac{1}{m}\sum_{i=1}^{m}(x_i-\mu_\beta)^2$$

The functions themselves are not relevant, I just included them for clarity. If I wanted to write out finding the derivative of $\frac{\partial{\hat{x_i}}}{\partial{\mu}}$, do I write it like this?

$$\frac{\partial{\hat{x_i}}}{\partial{\mu}} = \frac{\partial{\hat{x_i}}}{\partial{\mu}} + \frac{\partial{\hat{x_i}}}{\partial{\sigma^2}} \cdot \frac{\partial{\sigma^2}}{\partial{\mu}}$$

I feel like it looks confusing because I have $\frac{\partial{\hat{x_i}}}{\partial{\mu}}$ on both sides of the equal sign, so it could be difficult to interpret. Is there an alternative way to show that I'm trying to take the total derivative of the function as $\mu$ is in two arguments of $\hat{x_i}$?


1 Answer 1


I feel like it looks confusing because I have ∂xi^∂μ on both sides of the equal sign, so it could be difficult to interpret.

One is made with no independent variables constrained, the other made while constraining occurrences of $\sigma_\beta$ (treating it as though independent of $\mu_\beta$).

Too many authors leave this to context, but they should mark the distinction to avoid this confussion.

You may use the total derivative symbols, since that is what you are doing.

$$\large\dfrac{\mathrm d \hat x_i}{\mathrm d \mu_\beta}=\dfrac{\partial\hat x_i}{\partial \mu_\beta}+\dfrac{\partial\hat x_i}{\partial (\sigma_\beta^2)}\dfrac{\mathrm d(\sigma_\beta^2)}{\mathrm d\mu_\beta}$$

$\dfrac{\mathrm d \hat x_i}{\mathrm d \mu_\beta}$ is the total derivative of $\hat x_i$ with respect to $\mu_\beta$ (as the lone independent variable). Ie the derivative when no other independent variables are constrained.

$\dfrac{\partial\hat x_i}{\partial \mu_\beta}$ is the partial derivative of $\hat x_i$ with respect to $\mu_\beta$ while keeping $\sigma_\beta$ constrained.

Alternatively, you might indicating the set of independent variables being constrained.

$$\large\dfrac{\partial \hat x_i}{\partial \mu_\beta}=\left.\dfrac{\partial\hat x_i}{\partial \mu_\beta}\right\vert_{(\sigma_\beta^2)}+\left.\dfrac{\partial\hat x_i}{\partial (\sigma_\beta^2)}\right\vert_{\mu_\beta}\left.\dfrac{\partial(\sigma_\beta^2)}{\partial\mu_\beta}\right.$$

In this case, a partial derivative with no constrained independent variables is a total derivative.

You may also combine these approaches.

Alternatively you have functions $f( m, s), g(m)$ and define $\bar x_i=f(\mu_\beta, g(\mu_\beta))$ then::

$$\dfrac{\mathrm d\bar x_i}{\mathrm d\mu_\beta}= f_1(\mu_\beta, g(\mu_\beta))+f_2(\mu_\beta, g(\mu_\beta))\cdot g_1(\mu_\beta)$$

Where $f_1, f_2$ are partial derivatives of $f$ with respect to the first or second arguments, respectively.

  • $\begingroup$ Thanks for the different examples for notation! For the notation of the total derivative for your first example using $\frac{\mathrm d\hat{x_i}}{\mathrm d\mu_\beta}$, is this syntax identical to the univariate derivative? Is it expected for the reader to acknowledge it is the total derivative because I also show partials? e: I see the MathJax is a little small, I'm trying to refer to $\mathrm d$ $\endgroup$
    – Ryan
    Oct 6, 2021 at 0:24
  • $\begingroup$ Yes. When a function is univariate, the total derivative is identical to the ordinary derivative. In this case, $\bar x_i$ is presented as a bivariate function of $\mu_\beta$ and $\sigma_\beta$ but $\sigma_\beta$ itself is function of the former; so $\bar x_i$ is actually univariate. $\endgroup$ Oct 6, 2021 at 1:22

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