What does $\frac{dx}{dx}=1$ mean?

Question

What does it mean to differentiate a variable w.r.t itself and why such a thing exist in maths. I'm familliar with the mathematical explaination it's not what I'm looking for I find somehow a contradiction with the mathematical description and my logic, outside the mathematical context, if there's no change in any function wrt itself then then why would the $\frac{dx}{dx} = 1$ (which is something/quantity rather than nothing!), if I want to negate the existence of a change then wouldn't I simply say that $\frac{dx}{dx}$ is zero, zero change as the $\frac{d}{dx}$ (something) means the change that happens in that something wrt x

Answer 1

$$\frac{dx}{dx} \equiv \lim_{h \to 0} \frac{(x+h)-x}{h} = \lim_{h \to 0}\frac{h}{h} = 1$$

Answer 2

You have to read it this way: $$\left(\frac{d}{dx}\right)f(x),$$

so in your case it will be: $$\left(\frac{d}{dx}\right)x=1.$$

If you want to read it, it's the derivative of a function with respect to the variable $x$, and your function is $f(x)=x$.

Answer 3

It means differentiating the identity function $\operatorname{id}:x\mapsto x$ is derived for its argument $x$.

For a general picture, if you have $y=f(x)$ one way to write the derivative is $$\frac{\mathrm dy}{\mathrm dx} = f'(x).$$ Now for the identity function, $y=\operatorname{id}(x)=x$, and therefore $$\frac{\mathrm dx}{\mathrm dx} = \frac{\mathrm dy}{\mathrm dx} = \operatorname{id}'(x) = 1.$$

Edit for the updated question and comment:

The derivative tells you how much the value of the function changes if you change the argument of that function by a small amount. Indeed, that's where the notation comes from: $\frac{\mathrm dy}{\mathrm dx}$ is the quotient of the small change of the $y$ value ($\mathrm dy$) by the corresponding small change of the $x$-value ($\mathrm dx$). Now for the identity function, $y=x$, and therefore any change in $y$ is the very same as the change in $x$. That is $\mathrm dy=\mathrm dx$, and therefore $\frac{\mathrm dx}{\mathrm dx}=1$.

Answer 4

Why wouldn't you differentiate x with respect to itself. f(x) = x is a perfectly respectable function and we differentiate (or we try to) functions so we'd differentiate it. And why do you say "there is no change in the function with regards to itself"? There is plenty of change if f(x) = x. At x = 2, f(x) = 2. At x = 5, f(x) = 5. That's a change. What is its rate of change at change at x? Well, it's a constant rate of 1 unit change in x for 1 unit change in x. It's a constant rate of 1. A function that has no change is a constant function that never changes its value; f(x) = c. That has no change. So $\frac{dc}{dx}=0$.

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... or another way of putting it: you say that $\frac{d f(x)}{dx}$ is "change in the function with regards to x". Well, I'm not sure what you mean by that. A function defined as $f(x) = \sin x * \sqrt (\pi^x + x^{27.3}) + \frac {1}{x}^5 $ or whatever never changes to anything else with regards to x. It is always going to $ \sin x * \sqrt (\pi^x + x^{27.3}) + \frac {1}{x}^5 $ whatever x is.

What we are looking for is the rate of change in f(x) "with regards to" the rate of change in x (expressed in terms of x). And in that case the rate of change in x is exactly the same as the rate of change in x. If things are equal they are "1 in regard to each other". So $\frac{dx}{dx} =1$.

Answer 5

It also means that the slope of the line $y=x$ is 1.

Answer 6

Differentiation means "The rate of change of a variable with respect to other" or in simple terms how fast one variable is changing when we change another variable. its written as :

$\frac{d}{dx}f(x)$

In Geometry it means the slope of the function at a point on the graph of f(x)=y.

Thus if we differentiate $f(x)=x$ with respect to $x$ itself, we get $1$. That's because if we increase $x$ by 1, $f(x)$ also gets increased by $1$. Hence for every change in $x$, we get equivalent amount of change in $f(x)=x$. Hence the rate of change is always $1$.

Hope I didn't confuse you.

Answer 7

You say "if there's no change in any function wrt itself then then why[...]" but there is a change in the function with respect to $x$.

What you are trying to understand, always, when you take the derivative is how the value of the function changes when you change the value of the variable $x$.

When you take $x= 2$ then the value of the function is $2$ when you change the value of the variable $x$ from $2$ to $3$ then the value of the function also changes from $2$ to $3$.

The value of the function changes when you change the value of the variable. Thus the derivative is non-zero, and more specifically $1$ as the value of the function changes in exactly the same way as the variable.