Can 0^0 be defined when it expresses a gradient?

Question

According to power rule, ${d\over dx}(x) = 1x^0$. So, on the line $y=x$, we have ${dy\over dx}=x^0$. The gradient at any point on this line is 1. However, substituting 0 for x into the derivative of the function gives $0^0$, which is undefined. Does this mean the gradient at the point where the line passes through the origin is undefined?