I have been thinking about the second derivative from the Liebniz perspective.

Take the equation $y = x^3$.

The first derivative is $\frac{dy}{dx} = 3x^2$.

Now, the second derivative is commonly written like this:

$$\frac{d^2y}{dx^2} = 6x$$

However, if you actually perform the same operations to the left side that you did to the right, and said that $d^2u = d(d(u))$, then what you would actually have is:

$$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2} = 6x$$

This uses the quotient rule.

I have usually justified the reduction of the left side as being due to the fact that, if $\frac{d^2x}{dx^2}$ can be read as "the acceleration of x with respect to itself", then it should automatically reduce to zero, which would collapse the two definitions.

However, this doesn't really work. Because, if you followed this pattern further, we would be able to do the following:

$$\frac{d^2y}{dx^2}\frac{dx^2}{dy^2} = 6x \frac{dx^2}{dy^2} $$ $$\frac{d^2y}{dy^2} = 6x \frac{dx^2}{dy^2} $$ $$0 = 6x \frac{dx^2}{dy^2}$$

This is obviously wrong.

So, my question, then, is whether or not it is possible that the true second derivative is actually $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$, and perhaps $\frac{d^2x}{dx^2}$ goes to zero only in special (but common) cases (like when $x$ is the sole independent variable).

Note that I am working from an infinitesimal view of Calculus, that is, dx and dy refer to actual values (i.e., hyperreals).

Any thoughts on this?

(edited for clarity)

  • $\begingroup$ tha application of the quotient rule is meaningless. Leibniz symbol is not a quotient. $\endgroup$ – user Nov 27 '17 at 5:32
  • $\begingroup$ And Japan wasn't defeated in word war two. @gimusi $\endgroup$ – Mikhail Katz Jan 30 '18 at 13:12

In short, in the composition $y(x(z))$ where you intend to set $z=y$ in the end, the derivatives are per chain rule $$ \frac{dy}{dz}=\frac{dy}{dx}\,\frac{dx}{dz} $$ and $$ \frac{d^2y}{dz^2}=\frac{d^2y}{dx^2}\left(\frac{dx}{dz}\right)^2+\frac{dy}{dx}\,\frac{d^2x}{dz^2} $$ so that $$ 0=\frac{d^2y}{dy^2}=\frac{d^2y}{dx^2}\left(\frac{dx}{dy}\right)^2+\frac{dy}{dx}\,\frac{d^2x}{dy^2} $$ connects the second derivative of $y$ to the second derivative of the inverse function.


I have figured out that my original intuition about the second derivative seems to be correct - the second derivative of $y$ with respect to $x$ should be $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. More information can be found here:



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