chain rule problem of higher order derivatives

Question

I want to prove the following: $$\frac{\partial^2 y}{\partial s \partial s'}= \frac{\partial^2 x}{\partial s \partial s'}\frac{dy}{dx} + \frac{\partial x}{\partial s} \frac{\partial x}{\partial s'} \frac{d^2y}{dx^2}\\$$

My book derives it as follows:

Given $y = y(x)$ and $x = x(s, s')$ we have:

$$\frac{\partial y}{\partial s} = \frac{\partial x}{\partial s}\frac{dy}{dx}$$

by the usual chain rule. Differentiating with respect to $s'$ gives:

$$\frac{\partial^2 y}{\partial s \partial s'} \\ = \frac{\partial^2 x}{\partial s \partial s'}\frac{dy}{dx} + \frac{\partial x}{\partial s} \frac{d^2y}{\partial x \partial s'}\\ = \frac{\partial^2 x}{\partial s \partial s'}\frac{dy}{dx} + \frac{\partial x}{\partial s} \frac{\partial x}{\partial s'} \frac{d^2y}{dx^2}\\$$

However in the proof we considered: $$\frac{d^2y}{\partial x \partial s'}=\frac{\partial x}{\partial s'} \frac{d^2y}{dx^2}$$

How do we prove this?

Answer 1

What's the difference between $\frac{d}{dx}$ and $\frac{∂}{∂x}$? They seem to be used the same way."

Honestly, as far as I'm concerned, there's not really a difference between these operations. You could be pedantic and say one is only defined for single variable functions. But as far as intuition and computation go, they are one and the same and the difference is just meant to clarify what type of function is being differentiated.

Answer 2

To see that,first differentiate $y$ w.r.t. $s'$ and then again differentiate it w.r.t $x$. As the term $\frac{\partial x}{\partial s'}$ is independent of $x$ (it depends on $s$,$s'$ though) your desired result can be obtained.