1
$\begingroup$

I have the following function.

\begin{array}{l} \phi(w)=e^{b w}: \\ \frac{\partial \phi}{\partial t}=0, \quad \frac{\partial \phi}{\partial w}=b \phi, \quad \text { and } \quad \frac{\partial^{2} \phi}{\partial w^{2}}=b^{2} \phi \end{array}

I have trouble with calculating the derivative with respect to $w^2$. When I do reverse engineering with integrals, it makes sense. I would like to understand how to calculate this derivative. I find it difficult that you have to take the derivative with respect to $w^2$, but I don't see any $w^2$ in the expression.

Nadine

$\endgroup$
1
  • 2
    $\begingroup$ They are not taking the derivativ wrt $w^2$. You are differentiating wrt $w$ twice. $\partial^2 \phi/\partial w^2 = (\partial /\partial w)(\partial \phi/\partial w)$. $\endgroup$ – 0XLR Nov 29 '20 at 19:35
0
$\begingroup$

First of all, the standard notation for the second (and higher) derivatives is misleading, because it is non-algebraic (i.e., the ratios don't cancel). It's even more problematic when using partial differentials, because the symbols themselves are equivocal within the same expression.

In any case, what happens when you take a derivative is that you first take a differential and then divide by the differential of the variable you are taking the derivative "with respect to". When you take a derivative twice, you are dividing by that differential twice, which is why it is squared. Note that the proper interpretation of that notation is not $d(w^2)$ but rather $(d(w))^2$.

If you want an algebraically manipulable form of the second derivative, it is:

$$y'' = \frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2} = \frac{d(d(y))}{(d(x))^2} - \frac{d(y)}{d(x)}\frac{d(d(x))}{(d(x))^2}$$

I've expressed it above in two ways. The first is in more standard notation, the second is more explicit, so you can see what is being multiplied vs. what is being differentiated. Just note that the differential, in absence of parentheses, usually very tightly associates with the variable next to it.

So, the short answer, it is actually $(d(w))^2$. The long answer - if you are wanting something that is algebraically manipulable, and not mere symbology, you actually have to write it a different way. More information about this is available in my paper "Extending the Algebraic Manipulability of Differentials."

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.