# Function notation for the derivative operator

1. Jul 27, 2013

### cra18

If we have the function
$$f : x \mapsto f(x) = 3x^2,$$
I am used to Lagrange's prime notation for the derivative:
$$f' : x \mapsto f'(x) = 6x.$$
I'm fond of this notation. But it has been mostly abandoned in my engineering courses in favor of Leibniz's notation, using differential operators such as $\frac{\mathrm{d}}{\mathrm{d}x}$ per the above. I want to express the same relation as the second equation above using Leibniz's notation, but am having a lot of trouble. In particular, I understand that the derivative operator is a function that maps from a set of functions to a set of functions, so if
$$\frac{\mathrm{d}}{\mathrm{d}x} : f \mapsto \frac{\mathrm{d}}{\mathrm{d}x}f$$
is true, then it seems correct (to me) to write
$$\frac{\mathrm{d}}{\mathrm{d}x}f : a \mapsto \frac{\mathrm{d}}{\mathrm{d}x}f(a),$$
but I've been told that in order to show that a differentiated function is evaluated at some arbitrary point such as $a$, it is necessary to write something like
$$\frac{\mathrm{d}}{\mathrm{d}x}f(x)\vert_{x=a},$$
which means that the mappings as I have them are not correct, and it would be more correct to write something like
$$\frac{\mathrm{d}}{\mathrm{d}x} : f(x) \mapsto \frac{\mathrm{d}}{\mathrm{d}x}f(x)$$
and
$$\frac{\mathrm{d}}{\mathrm{d}x}f(x) : a \mapsto \frac{\mathrm{d}}{\mathrm{d}x}f(x)\vert_{x=a}.$$

Any help would be greatly appreciated; I'm starting to confuse myself. I am just trying to express the derivative of a function using Leibniz's notation in arrow notation.

2. Jul 27, 2013

### Staff: Mentor

That's difficult to believe. Prime notation is usually more compact than Leibniz notation, if somewhat less explicit, and engineers tend to strive for efficiency.
No, this isn't correct, since d/dx(f(a)) = 0. f(a) is a constant, so its derivative is zero.
This seems very cumbersome.

3. Jul 27, 2013

### cra18

You're right. It would have to be something like
$$\frac{\mathrm{d}}{\mathrm{d}x}f : a \mapsto \left(\frac{\mathrm{d}}{\mathrm{d}x}f\right)(a) = \frac{\mathrm{d}}{\mathrm{d}x}f(x)\vert_{x=a}.$$
This is indeed extremely cumbersome, which is probably why I haven't been able to find an example doing this. I think I'll stick with the definition
$$f' : x \mapsto f'(x) = \frac{\mathrm{d}}{\mathrm{d}x}f(x),$$
though I hate mixing notation.

Thanks for your insight. And the reason why Leibniz notation is favored here is because of how easily linear operators can be implemented in Matlab or some such language using symbolic math packages.

Last edited: Jul 27, 2013
4. Jul 27, 2013

### jbunniii

There are other notations for the differentiation operator which aren't saddled with the cumbersome $x$ of $d/dx$. For functions of a single variable, a simple $D$ is often used:
$$D : f \mapsto f'$$
This notation has the advantage that iteration of the operator can be conveniently indicated with an exponent:
$$D^n : f \mapsto f^{(n)}$$
For functions of multiple variables, one sometimes sees notation like $D_k$ to indicate partial differentiation with respect to the $k$'th variable.

5. Jul 27, 2013

### lurflurf

It is important to know all of the common notations, both because they are in common use and to switch to the form that best in a particular situation. We indicate the

$$\left( \dfrac{d }{dx} \right)^n f(x)=\dfrac{d^n }{dx^n} f(x)=\dfrac{d^n f(x)}{dx^n} = (f(x))^{(n)}=D^n f(x) =D_x^n f(x)$$
The trouble is that f(a) has derivative zero so we must indicate we want to take the derivative, then evaluate it. Not evaluate it, then take the derivative. Two ways to indicate this are to use the evaluation bar, or use the derivative notation without the variable and evaluate the indicated derivative
$$\left. \left( \dfrac{d }{dx} \right)^n f(x) \right| _{x=a} =\left. \dfrac{d^n }{dx^n} f(x) \right| _{x=a} =\left. \dfrac{d^n f(x)}{dx^n} \right| _{x=a} = \left. (f(x))^{(n)} \right| _{x=a} = \left. D^n f(x) \right| _{x=a} = \left. D_x^n f(x) \right| _{x=a} \\ \\ \left( \left( \dfrac{d }{dx} \right)^n f \right) (a) =\left( \dfrac{d^n }{dx^n} f \right) (a) =\left( \dfrac{d^n f}{dx^n} \right) (a) = f^{(n)} (a) =(D^n f)(a) =(D_x^n f)(a)$$

in summary

$$D(f(x))=(D f)(x)=\dfrac{d}{dx} f(x) \\ (D f)(a)=\left. \dfrac{d}{dx} f(x) \right| _{x=a} \\ D(f(a))=\dfrac{d}{dx} f(a)=0$$

6. Jul 27, 2013

### cra18

Thanks for all of the answers. And that was incredibly helpful lurflurf.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook