Is $c \frac{\partial}{\partial x}=\frac{\partial c}{\partial x}=0$?

Question

Is $c \frac{\partial}{\partial x}=\frac{\partial c}{\partial x}=0$?

Or is one supposed to treat $c \frac{\partial}{\partial x}$ as "partially applied" operator (which expects a function).

Answer 1

If c is a number , yeah thats right We know that Differentiate of every number relative to x is equal to 0 so the second side is equal to 0 . (differentiate of c and 1 relative to x is 0) so in the left side we have : [c(0)=0] So yeah the equation is true.

Attention: my original language is not english and im not proffesional on it . so excuse me if i have mistake in my writing or in my words)

Answer 2

Let's have your 2 operators:

• $\hat{D}=\frac{\partial}{\partial x}$

• $\hat{C}=c$ -yes, in the theory can be as well

Now test on a function $f(x)$- $\hat{C}(\hat{D}f(x))$ and $\hat{D}(\hat{C}f(x))$:

1) $\hat{C}(\hat{D}f(x))= \hat{C}(\frac{\partial}{\partial x}f(x))=\hat{C}f'(x)=c\cdot f'(x)$

2) $$\hat{D}(\hat{C}f(x))=\hat{D}(c\cdot f(x))=\frac{\partial}{\partial x}(c\cdot f(x))=[\frac{\partial}{\partial x}c]\cdot f(x) +c\cdot [\frac{\partial}{\partial x}f(x)]=0\cdot f(x)+c\cdot f'(x)=c\cdot f'(x)$$

True is here $\hat{C}\cdot\hat{D}=\hat{D}\cdot\hat{C}$ but not $c\cdot\hat{D}=\hat{D}\cdot c$

Answer 3

I don't see how "linearity" of the derivative, that $\frac{\partial}{\partial x}(af(x)+ bg(x))= a\frac{\partial f}{\partial x}+ b\frac{\partial g}{\partial x}$, has anything to do with this. It is just a matter of notation- understanding what $c\frac{\partial}{\partial x}$ means as differential operator.