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Derivative notation

MacLaddy

Member
Jan 29, 2012
52
I am always getting mixed up on derivative notation, so I was just wondering if this below makes sense?

\(f(x) = 2x^\sqrt{2}\)

\(\frac{df(x)}{dx} = 2\frac{d}{dx}x^\sqrt{2}\)

The first should probably just be \(\frac{dy}{dx}\), but I was wondering if the other way would work as well.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
I am always getting mixed up on derivative notation, so I was just wondering if this below makes sense?

\(f(x) = 2x^\sqrt{2}\)

\(\frac{df(x)}{dx} = 2\frac{d}{dx}x^\sqrt{2}\)

The first should probably just be \(\frac{dy}{dx}\), but I was wondering if the other way would work as well.
Actually, the way you have it is perfectly fine, and better than $dy/dx$, unless you've defined $y=f(x)$. Another equally valid notation is $f'(x)$.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I am always getting mixed up on derivative notation, so I was just wondering if this below makes sense?

\(f(x) = 2x^\sqrt{2}\)

\(\frac{df(x)}{dx} = 2\frac{d}{dx}x^\sqrt{2}\)

The first should probably just be \(\frac{dy}{dx}\), but I was wondering if the other way would work as well.
Using the fact $y=f(x)$ Then you can write $\dfrac{df(x)}{dx}=\dfrac{dy}{dx}$

And yes you can do the above.
 

MacLaddy

Member
Jan 29, 2012
52
Thanks Ackbach and dwsmith. I've never seen my instructor do it that way, but it seemed to make sense.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
If $y=f(x)$, the following are all equivalent:

$$Dy=Df(x)=\frac{d}{dx}\,y=\frac{d}{dx}\,f(x)=y'=f'(x)=\frac{dy}{dx}=\frac{df(x)}{dx}.$$

And I'm probably leaving out a few notations. Hope this doesn't confuse you, but this is the way it's developed.