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I have the function $f(X)$, $X$: depend to a variable $\theta $. is there a direct expression to calculate the derivative of $f(X)$ with respect to $\theta$:

$$\frac{df(X)}{d\theta} $$

And I want to provide me with some references if you would.

user137684
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1 Answers1

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According to the chain rule:

$$\frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta},$$

that is, you first differentiate the outer function $f$ wrt. to the inner function $x$, and then differentiate the inner function wrt. to the variable you're originally differentiating wrt. (here $\theta$), and then you multiply them to get the desired result.

EDIT, as response to OPs edit:

It seems you are looking for the so-called total derivative, where a hard d is used:

$$\frac{\mathrm{d} f}{\mathrm{d} \theta} = \frac{\partial f}{\partial \theta}+\frac{\partial f}{\partial x} \frac{\partial x}{\partial \theta}.$$

Note that $$\frac{\mathrm{d} x}{\mathrm{d} \theta}=\frac{\partial x}{\partial \theta},$$ as $x$ only depends on $\theta$ (at least, this is assumed from the information given).

Bobson Dugnutt
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  • I found some references that they use this expression : $\frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial X} \frac{\partial X}{\partial \theta} + \frac{\partial f}{\partial \theta} $, Is it true and What is the difference between it and your expression. – user137684 Feb 07 '17 at 10:35
  • @user137684, just superficially that expression does not make sense, because you ought to be able to cancel $\frac{\partial f}{\partial \theta}$ on both sides, ensuring that the other term is always 0. Are you sure that's what you read? – Mees de Vries Feb 07 '17 at 10:40
  • Sorry I made I mistake (difference between $d$ and $\partial$), and I updated my post $\frac{d f}{d \theta} = \frac{\partial f}{\partial X} \frac{d X}{d \theta} + \frac{\partial f}{\partial \theta}$. – user137684 Feb 07 '17 at 10:51
  • @user137684 See my update. – Bobson Dugnutt Feb 07 '17 at 16:40
  • @ Lovsovs, Thanks for your answer. Last thing I think in right hand side of your updated expression first part, $\frac{\partial f}{\partial \theta}$ is the correct. – user137684 Feb 08 '17 at 08:21
  • @user137684 Yes of course, thank you! :) – Bobson Dugnutt Feb 08 '17 at 13:52
  • @user137684 If you liked the answer, you can upvote and/or accept it. If it didn't answer your question, please do let me know where it left you wanting. Cheers. – Bobson Dugnutt Feb 11 '17 at 20:37