MHB Derivative Notation: Clarifying Confusion

[MacLaddy1]

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]

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 $f'(x)$.

 

[Dustinsfl]

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.

 

[MacLaddy1]

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

 

[Ackbach]

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.