# How is the derivative calculated with respect to w^2?

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.

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

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