Why is the differentiated 2x 2 but not x squared?

[ryanuser]

I apologise because my question is more mathematically related than physical, however I was unsuccessful at finding a better place to ask. My question is how the differentiated 2x is 2 but not x squared? I learned that In order to differentiated, we place the number that x has been powered to, in front of x then subtract the number from 1 to get the new power of x.
Thus I thought x to power of 2 is the answer as we swapped the 2 by the x's power.
Have I misunderstood the concept of differentiation or i am missing something here?

Thanks

 

[Orodruin]

It is not about swapping the exponent and coefficient. In general, the derivative of x^n is nx^(n-1). The derivative is also a linear operation, so if you multiply a function by a constant, the derivative of the function is also multiplied by that constant.

 

[Matterwave]

There is a calculus help forum here, it is like 2 forums down. To answer your question, we can look directly at the definition of differentiation:

$$f'(x)\equiv \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$

If $f(x)=2x$ then we see:

$$f'(x)= \lim_{h\rightarrow 0}\frac{2(x+h)-2x}{h}=\lim_{h\rightarrow 0}\frac{2h}{h}=2$$

More generally, as Orodruin mentioned, the derivative of $x^n$ is equal to $nx^{n-1}$. In your case, n=1 so that the derivative of $x$ is simply $1$.

Lastly, the derivative is linear so that for some constant $c$:

$$\frac{d}{dx}(cf(x))=c\frac{d}{dx}f(x)$$

So, combining our knowledge here we find again:

$$\frac{d}{dx}2x=2\frac{d}{dx}x=2\cdot 1=2$$

This matches what we found by directly appealing to the definition of the derivative.

 

[ryanuser]

Thanks for both of your explanations, now it became more clear to me.

 

[thelema418]

Beside the rigorous definition, remember also the connection of the derivative to the concept of slope and tangent lines.

From algebra, you know the form y = mx + b is the equation of a line with slope m for every x value. So if f(x) = 2x, the derivative is the slope of the line, which is 2.

 

[epenguin]

Your idea in #1 is a garbled version vaguely resembling #2 but wrong. You see it looks like you have relied on memory.

I do not say that is wrong or avoidable. I am sure we all do it a lot, and it might take a bit of effort if we had to produce proofs of some of the things we reply on, but if whenever we find have that difficulty it is a good practice to do some revision and we then benefit from increasing clarity and fixing things in mind.

And connect things up. Actually it is the integral of 2x with respect to x that is x². (the differentiated x² is 2x). You might have confused with that. You can easily see both this result and the one of your question graphically.