# Derivative Notation Issue

• Dec 28th 2010, 08:26 PM
Skyrim
Derivative Notation Issue
Guys I am sorry if this question has been asked before, but it has been pestering me for long time. It is about the notation of derivatives using 'prime'.

If f(x) is a function of x, then does f'(a) represent f(a) differentiated with respect to a, or does it represent f(x) differentiated with respect to x where x=a or are they both the same thing?

Forgive the lack of Latex sophistication.
• Dec 28th 2010, 08:35 PM
TheCoffeeMachine
Quote:

Originally Posted by Skyrim
If f(x) is a function of x, then does f'(a) represent f(a) differentiated with respect to a, or does it represent f(x) differentiated with respect to x where x=a or are they both the same thing?

If your function is f(x), then f'(x) is the derivative of f(x); and f'(a) is the derivative of f(x) evaluated at x = a. If your function is f(a), however, then f'(a) of course represents the derivative of f(a).
• Dec 29th 2010, 01:22 AM
HallsofIvy
In other words, it depends upon whether you are thinking of "a" as a variable or as a specific value of a variable. If your function has been written as a function of x and then you see f'(a), that is the derivative with respect to the variable x, evaluated at x= a.

Of course, if you think of a as being the variable, and differentiate f(a) with respect to a, you should get exactly the same thing as if you differentiated f(x) with respect to x, then set x= a.

For example, if the function is $\displaystyle f(x)= x^3- sin(x)$ then $\displaystyle f'(x)= 3x^2- cos(x)$ and, setting x= a, $\displaystyle f'(a)= 3a^2- cos(a)$. On the other hand, if I replace the variable x with the variable a, I would have $\displaystyle f(a)= a^3- sin(a)$ and the derivative with respect to a is $\displaystyle f'(a)= 3a^2- cos(a)$, exactly as before.

Of course, that has to be a letter or something that we can think of as a variable. If we have $\displaystyle f(x)= x^3- sin(x)$ as before and are asked to find $\displaystyle f'(\pi)$, we would set $\displaystyle f'(x)= 3x^2- cos(x)$ and then set $\displaystyle x= \pi$: $\displaystyle f'(\pi)= 3\pi^2+ 1$.

It would be a terrible mistake to first set x equal to the constant $\displaystyle \pi$ and then argue "$\displaystyle f(\pi)= \pi^3- sin(\pi)= \pi^3$ is a constant so its derivative is 0". We differentiate with respect to a variable, not a constant.
• Dec 29th 2010, 06:52 AM
Skyrim
Quote:

Originally Posted by HallsofIvy
In other words, it depends upon whether you are thinking of "a" as a variable or as a specific value of a variable. If your function has been written as a function of x and then you see f'(a), that is the derivative with respect to the variable x, evaluated at x= a.

Of course, if you think of a as being the variable, and differentiate f(a) with respect to a, you should get exactly the same thing as if you differentiated f(x) with respect to x, then set x= a.

For example, if the function is $\displaystyle f(x)= x^3- sin(x)$ then $\displaystyle f'(x)= 3x^2- cos(x)$ and, setting x= a, $\displaystyle f'(a)= 3a^2- cos(a)$. On the other hand, if I replace the variable x with the variable a, I would have $\displaystyle f(a)= a^3- sin(a)$ and the derivative with respect to a is $\displaystyle f'(a)= 3a^2- cos(a)$, exactly as before.

Of course, that has to be a letter or something that we can think of as a variable. If we have $\displaystyle f(x)= x^3- sin(x)$ as before and are asked to find $\displaystyle f'(\pi)$, we would set $\displaystyle f'(x)= 3x^2- cos(x)$ and then set $\displaystyle x= \pi$: $\displaystyle f'(\pi)= 3\pi^2+ 1$.

It would be a terrible mistake to first set x equal to the constant $\displaystyle \pi$ and then argue "$\displaystyle f(\pi)= \pi^3- sin(\pi)= \pi^3$ is a constant so its derivative is 0". We differentiate with respect to a variable, not a constant.

Thank you so much.