Is this generalisation of chain rule true?

Question

Is the following equation true:(Here y is a function of x & x is a function of s and s') $$\dfrac{dy}{dx}\dfrac{d^{2}x}{dsds'}=\dfrac{d^{2}y}{dsds'}$$ If it is true, please give an easy to understand proof.

Answer 1

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:

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

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A quick example (as a sanity check): Take $y = \sin x$, $x = s^2 s'$. We calculate directly that:

$$\frac{\partial^2 y}{\partial s \partial s'} = 2s \cos (s^2 s') - 2s^3 s' \sin(s^2 s')$$

On the other hand, we have: $$\frac{\partial^2 x}{\partial s \partial s'} = 2s \hspace{1cm} \frac{dy}{dx} = \cos x$$ $$\frac{\partial x}{\partial s} = 2s s' \hspace{1cm} \frac{\partial x}{\partial s'} = s^2 \hspace{1cm} \frac{d^2 y}{dx^2} = -\sin x$$

Thus our formula on the RHS becomes: $$2s \cos(s^2 s') + 2s s' \cdot s^2 \cdot - sin(s^2 s') = 2s \cos (s^2 s') - 2s^3 s' \sin(s^2 s')$$

as desired.