# Differentiating $x=1$ with respect to $x$

Sorry this may sound like a silly question and I know that this does't meet the quality standards of Math S.E, I found this in one of the Math-Jokes websites and found it interesting,

$$x=1$$ $$\frac{d}{dx} x=\frac{d}{dx} 1$$ $$1=0$$

What I think is the problem is that it is not differentiable, saying $x=1$ is equivalent to saying $y=\delta(x-1)$, which as we know is not continuous at $x=1$, and so it is not differentiable. is this right? or is there some other reason to resolve the paradox?

Define functions $f(x) = x$ and $g(x) = 1$, where both $f$ and $g$ have a domain which includes the point $p = 1$, say the interval $J = (1-\delta,1 + \delta)$ for some $\delta > 0$.

Then clearly $f$ and $g$ are not identical on $J$, nor do they vary by a constant on all of $J$. Hence there is no reason to expect their derivative will be equal on $J$.

The only place where $f$ and $g$ are equal is the point $p = 1$. However, that is neither a necessary or sufficient condition for the derivative of $f$ or $g$ to be equal at $p$. For instance

• the functions $h_1(x) = 2, h_2(x) = 3$ are equal no where on all of the reals but their derivatives are equal everywhere
• the functions $h_3(x) = 0, h_4(x) = \sin x$ are equal infinitely often on the domain of all the reals but their derivatives are never the same on that set of points

The way to look at this is that we do not have a function which is what differentiation takes as its input.

Also, remember a derivative can be thought of as the slope of a function as the distance between two points as that distance approaches zero. But no function no slope. If the equation was : $$x(y)=1$$ which is defined on the line $x=1$, then you could take the derivative with respect to $y$ and get $$x\prime = 0$$

Basically the definition of a derivative which is:$$\lim{x\to0} \ \frac{{f(x + \Delta x ) - f\left( x \right)}}{\Delta x}$$

as you can see it requires a function and the equation $x =1$ is not so differentiating becomes an invalid operation!

There should be two distinctly different variables to have a meaning in differentiation.

$y=1, \dfrac{dy}{dx} = 0.$