# Derivative notation

##### Member
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
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

#### dwsmith

##### Well-known member
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.

##### Member
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
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.

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