# Does an undefined derivative always mean a vertical tangent line, and why do we define a tangent line when the derivative is undefined?

I am wondering whether an undefined derivative at a point implies that the tangent line to that point is vertical, and also how the tangent line could still exist if the derivative doesn't exist. For my first question, although I have read that vertical tangent lines will have an undefined derivative, I am wondering whether every single instance of a derivative being undefined at a point means that the tangent line at that point is vertical, and that there isn't another explanation for an undefined derivative. I am confused on this because my Professor intermediately concludes a vertical tangent line when the derivative is undefined, but it seems any situation where the derivative doesn't exist would lead to an undefined derivative, which may not necessarily have a vertical tangent line.

For my second question, what is the reason why we regard the tangent line as existing when the derivative doesn't exist when we define the tangent line to a point to have slope equal to the derivative at that point? It seems that if a component of the tangent (its slope) doesn't exist, then the tangent line itself wouldn't exist.

UPDATE: I am now wondering whether the derivative being of the form $$a \over 0$$ implies that there is a vertical tangent line after reading the answers. Can a derivative of the form $$a \over 0$$ eventualize for any other reason that a vertical tangent line?

• No, a point of discontinuity need not have any tangent. Commented Jul 26, 2023 at 17:38
• Same for a point of non-differentiability where the function is continuous. Commented Jul 26, 2023 at 17:39
• @RobertIsrael if the reason the derivative doesn't exist is because it is of the form $a \over 0$, then could we conclude a vertical tangent line? Commented Jul 26, 2023 at 17:41
• It doesn't imply that. It only implies this if $\lim_{h\to0}\left|\frac{f(x+h)-f(x)}{h}\right|$ converges to $\infty$ and $\lim(f(x+h)-f(x))=0.$ Commented Jul 26, 2023 at 18:12