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Notation for derivatives

  • Thread starter dranseth
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1. Homework Statement

I cannot seem to find my textbook and we just started derivatives. Can anyone tell me when I would use the notation dy/dx as opposed to something like f'(x)??

Thanks!
 

Answers and Replies

334
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f'(x) is just function notation. If you're given an f(x), it's equally correct to use f'(x) or d(f(x))/dx, for example. In the case of f'(x), the derivative with respect to x is implied.
 
88
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So it wouldn't ever matter which one I use? I could use both?
 
334
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You should stick with one for consistency. There's nothing wrong with using y' over dy/dx, or f '(x) over d(f(x))/dx; Just be consistent.
 
rock.freak667
Homework Helper
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So it wouldn't ever matter which one I use? I could use both?
Not loosely in a sense.

For example, if you have

[itex]y=x^2[/itex]
You would write [itex]\frac{dy}{dx}[/itex] or [itex]y'[/itex] = [itex]2x[/itex] and NOT [itex]f'(x)=2x[/itex]
 
nicksauce
Science Advisor
Homework Helper
1,272
5
The dy/dx notation is nice when you're actually manipulating these as if they're actually numbers. It's a physicist's favorite trick (but a mathematician's worst nightmare). I for one am not a fan of f'(x), except when there are higher order derivatives involved.
 
I'd get used to both forms, as they seem to turn up alternately, at least in my experience.

For example the quotient rule:

Leibniz notation:

[tex]\frac{d}{dx}\left (\frac{u}{v}\right ) = \frac{\frac{du}{dx}\cdot v-u\cdot\frac{dv}{dx}}{v^2}[/tex]

Newtonian notation:

[tex]h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}[/tex]

or in shorthand notation:

[tex]h'=\frac {f'\cdot g-f\cdot g'}{g^2}[/tex]

You might see something similar as well which is a Newtonian notation:

[tex]\dot{h}(x)=\dot{f}(x)+\dot{g}(x) \longrightarrow \ddot{h}(x)=\ddot{f}(x)+\ddot{g}(x)[/tex]

or

[tex]h\,\dot{}\,(x)=f\,\dot{}\,(x)+g\,\dot{}\,(x) \longrightarrow h\,\ddot{}\,(x) = f\,\ddot{}\,(x) + g\,\ddot{}\,(x)[/tex]

occasionally but these aren't used very often.

Which is the same as saying:

[tex]\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}\longrightarrow \frac{d^2}{dx^2}(u + v)=\frac{d^2u}{dx^2}+\frac{d^2v}{dx^2}[/tex]

in Leibniz notation.

They look more complicated between forms sometimes, but once you get the idea they're pretty obvious.
 
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