4
$\begingroup$

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?

$\endgroup$

3 Answers 3

8
$\begingroup$

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
$\endgroup$
4
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .