I know that $|x|$ is not differentiable at $x=0$ because there is potentially an infinite number of tangent lines going through that point.

But let's say we were interested in the motion of an object through time starting at the point $0$ and going forward, wouldn't we then be able to talk about the instantaneous rate of change of $y$ with regard to $x$ at $x=0$? And wouldn't it be 1?

If not, how would we be able to talk about the rate of change of position to time at the point $0$?

I'm having trouble reconciling those two ideas; namely that there seems to be a rate of change at $0$ if we specify a direction on the $x$ axis, while at the same time the derivative technically doesn't exist at that point.

  • 3
    $\begingroup$ There is a right derivative (it's $1$), and a left-derivative ($-1$) at $0$, yes. See e.g. this. $\endgroup$ – Clement C. Jun 17 '16 at 3:13
  • $\begingroup$ @ClementC. Are those considered directional derivatives like the ones in multivariable calculus? $\endgroup$ – jeremy radcliff Jun 17 '16 at 3:14
  • 2
    $\begingroup$ No, it's not the same... for directional derivatives, you look at a direction $\vec{v}$ and consider $\frac{f(a+t\vec{v})-f(a)}{t}$, with no restriction on the sign of $t$. Here, you do restrict the sign. $\endgroup$ – Clement C. Jun 17 '16 at 3:16
  • $\begingroup$ @ClementC. Thank you for clarifying. $\endgroup$ – jeremy radcliff Jun 17 '16 at 3:23

If the domain of $\left |x \right |$ is restricted to +R surely the derivative at x=0 exists and the value is equal to 1.
And similarly if we restrict domain to -R the value is -1.
This is what you tried to explain by saying that you started moving in +x direction,this is nothing but domain restriction.
Assume that we haven't restricted the domain then there would have been a clash between the direction of velocity as I hve mentioned above and hence the fuction is not differentiable at x=0.

  • $\begingroup$ But isn't -R=+R? $\endgroup$ – 355durch113 Jun 17 '16 at 4:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.