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Was just going through my single variable calculus notes recently when I came across this interesting article on the relation between differentiabilty of a function. I missed some of the points my professor made and hence a few doubts remain in my mind -

  1. Can a function be differentiable at a point and still have no tangent at that point?

  2. Can a function have a tangent at a point and still be non-differentiable at the point?

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Don't mix functions with their graph. The tangent line to the graph of a funtion has slope which is the derivative of the function. –  Emanuele Paolini Mar 12 at 20:10
    
We all know what a derivative is. But what is a tangent? –  Christian Blatter Mar 12 at 20:53

2 Answers 2

up vote 2 down vote accepted

Differentiability at a point, for a real-valued function of one variable, is the same as the existence of a tangent line at that point, except for one case: If the tangent line is vertical, then the function is not differentiable at that point. It's because a vertical line isn't represented by any linear function, and differentiability is really about looking locally like a linear function.

Thus, the answer to your first question is "no", and the answer to your second question is "yes, consider $y=x^{1/3}$ at $(0,0)$"

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Thanks! The second point you made about $y=x^{1/3}$ - so is it correct to conclude that if a function has a unique tangent at point, it'll be differentiable at that point? –  Mathguy Mar 12 at 20:00
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No, the cube root function has a unique tangent line at the origin, namely the y-axis. What's required is a non-vertical tangent line. –  G Tony Jacobs Mar 12 at 20:02
    
So is it a necessary condition that the tangent be unique for the function to be differentiable? For instance, for $y=|x|$, the tangent at origin - does it exist? If yes, then is it unique? –  Mathguy Mar 12 at 20:05
    
@Mathguy the tangent being unique is necessary but not sufficient. –  Sabyasachi Mar 12 at 20:06
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The necessary and sufficient condition is the tangent must be unique and be non-vertical –  Sabyasachi Mar 12 at 20:07

If a function is differentiable at some point $(x_0,y_0)$ with slope $m$, then the tangent is $y-y_0=m(x-x_0)$. And so, the first proposition is false.

As for the second proposition, it is true, yes a function can have a tangent without being differentiable. Consider the function $$y=\sqrt{25-x^2}$$

It has tangents $x=5$ and $x=-5$ but it is not differntiable at these points of tangency.

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