1

While writing the second derivative of y, $\frac{d^2y}{dx^2}$ what does the symbol $dx^2$ signify? I know that in case of the first derivative $dy$ means change in y and $dx$ means change in y and also that $d^2$ means change of change of y.

Soham
  • 2,029
  • This is not really a very nice notation in modern standards. $dx^2$ or $dxdx$ is a symmetric order-2 tensor, which usually appears in differential geometry. But here, like grandpa said, it does not have a meaning on its own. You could probably find O. Loos's little book on symmetric space helpful. In his chapter one, he explained various sorts of differentiation using the unified language of tensor. – Troy Woo Sep 15 '14 at 06:41

1 Answers1

3

It does not have a meaning on its own. $\dfrac{d^2 y}{dx^2}$ is just an abbreviation for $\dfrac{d \left(dy/dx\right)}{dx}$, that is the derivative of $\dfrac{dy}{dx}$.

Robert Israel
  • 448,999
  • So it's like a shorthand for multiplication $\dfrac{d \left(dy/dx\right)}{dx} = \dfrac{d(dy)}{dx \cdot dx} = \dfrac{d^2 y}{(dx)^2}$, even though one cannot really do this kind of operations with differentials? – rubik Sep 15 '14 at 06:55
  • 1
    Just use the f(x + E) = f(x) + E.f'(x) notation. dy/dx has always been confusing. –  Sep 15 '14 at 07:09