If we took $y = x^2$ and cut it in half by letting $x\ge 0$, does the derivative still exist at $x = 0$ or is it $\text{DNE}$?
I think it's still $0$ because the function and it's derivative are both still continuous. However, my friends have told me that it's $\text{DNE}$ for the simple reason that the left and right limits at $x = 0$ don't exist. I don't understand this logic.
Is $y = x^2, x \ge 0$ differentiable at $x = 0$?
when you cut off the parabola, you cut off half the tangent at $x = 0, y = 0$ with it. so what remains is half the tangent. there is definition of one sided derivative at the boundaries of the domain. for example, at the left boundary $x = a,$ the one sided derivative of $f$ is defined by $$D_+f|_{x = a} = \lim_{x \to a+} = \frac{f(x) - f(a)}{x-a}.$$
likewise, you can define $D_-f|_{x = b}$ at the right end point $b.$