# Is there a derivative for $|x|$ at $0$ specifically “in the direction” of positive $x$?

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.

• There is a right derivative (it's $1$), and a left-derivative ($-1$) at $0$, yes. See e.g. this. – Clement C. Jun 17 '16 at 3:13
• @ClementC. Are those considered directional derivatives like the ones in multivariable calculus? – jeremy radcliff Jun 17 '16 at 3:14
• 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. – Clement C. Jun 17 '16 at 3:16
• @ClementC. Thank you for clarifying. – 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.