Properties of Differential Operator and Proper Formalism

[Nano-Passion]

Let us take a function defined by

$$y=cx$$
To differentiate that, we use the operator d/dx
$$\frac{d}{dx} y = \frac{d}{dx} cx^{-1}$$
By the chain rule/implicit differentiation on the left and normal differentiation on the right we get,
$$\frac{dy}{dx} = -1c x^{-2}$$

What confuses me is the proper usage of the formalism to get

$$\frac{d^2y}{dx^2} = 2cx^{-3}$$

It doesn't seem that we can use the same operation as last time.

 

[meldraft]

Your initial calculation is wrong, x is not supposed to be x^(-1), it remains x, and after the differentiation it is 1. The second derivative is then, of course, 0.

 

[AlephZero]

Why not?

$\dfrac{d^2y}{dx^2}$ is just a (slightly confusing, if you think too hard about what $dx^2$ is supposed to mean) abbreviation for $\dfrac{d}{dx}\left(\dfrac{d}{dx}y\right)$.

(I assumed your first equation was a typo for $y = cx^{-1}$).

 

[Nano-Passion]

Your initial calculation is wrong, x is not supposed to be x^(-1), it remains x, and after the differentiation it is 1. The second derivative is then, of course, 0.

Typo.

Why not?

$\dfrac{d^2y}{dx^2}$ is just a (slightly confusing, if you think too hard about what $dx^2$ is supposed to mean) abbreviation for $\dfrac{d}{dx}\left(\dfrac{d}{dx}y\right)$.

(I assumed your first equation was a typo for $y = cx^{-1}$).

Oh thank you! Now I realize why it is written in that form