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Can anyone help me out? Why every point of a function where differentiation exists has only one tangent? I know the slope at any point of any function is defined by differentiation at that point.But there may be another straight line which touches the function at that point but slope is different from the differentiation value at that point. But this does not happen(I am not considering singular point).

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    $\begingroup$ How d you define "touch"? $\endgroup$ – Hagen von Eitzen Jul 1 '16 at 16:06
  • $\begingroup$ Touch means what u r thinking.It will have one point common(let's say C) with the curve and there is an interval around that point in where the straight line has no common point other than C with the curve. $\endgroup$ – sani Jul 1 '16 at 16:08
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    $\begingroup$ Say $f(x)=x$. Then, in your sense, every line of the form $y=mx$ with $m\neq 1$ touches the graph of $f(x)$ at $(0,0)$ (and, ironically, the tangent to the curve at that point does not touch the graph in your sense). $\endgroup$ – lulu Jul 1 '16 at 16:15
  • $\begingroup$ yea I am wrong..What is ur definition $\endgroup$ – sani Jul 1 '16 at 16:20
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    $\begingroup$ Take a look at my answer to this question. It may help you make sense of this. $\endgroup$ – user137731 Jul 1 '16 at 16:53
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It's not just about a line touching. (Every line through the point touches the graph there.) It's the best linear approximation, in the sense that the error $f(x)-L(x)$ goes to $0$ faster than $x$ approaches $a$. (Here $L(x)$ is the linear function whose graph we're discussing.)

EDIT: To be more precise, if $f$ is differentiable at $x=a$, then the linear function $L(x)=f(a)+f'(a)(x-a)$ is the unique linear function with the property that $$\lim_{x\to a}\frac{f(x)-L(x)}{x-a} = 0.$$ Note that if we take $L(x)=f(a)+m(x-a)$, this limit becomes $$\lim_{x\to a}\frac{f(x)-f(a)-m(x-a)}{x-a} = \lim_{x\to a}\frac{f(x)-f(a)}{x-a} -m = f'(a)-m,$$ and this is $0$ if and only if $m=f'(a)$.

If $f$ fails to be differentiable, this limit will never be 0. Try it, for example, with $f(x)=|x|$ at $x=0$. If you take a line of slope $m$ and $x>0$, this limit is $\lim_{x\to 0^+} \dfrac{x-mx}x = 1-m$, so $m$ would have to be $1$, but if $x<0$, this limit is $\lim_{x\to 0^-} \dfrac{-x-mx}x = -1-m$, so $m$ would have to be $-1$. There therefore is no tangent line at $x=0$.

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  • $\begingroup$ NO no if differentiation exists at that point. Only one straight line would touch the curve at that point. Touch means what I have said earlier in a comment $\endgroup$ – sani Jul 1 '16 at 16:12
  • $\begingroup$ Forgive me I want to ask you then why there is no derivative at sharp point of a curve? $\endgroup$ – sani Jul 1 '16 at 16:43
  • $\begingroup$ @sani: See my edit. I hope it helps. $\endgroup$ – Ted Shifrin Jul 1 '16 at 17:18
  • $\begingroup$ Thank you @Ted...It really helped me.. $\endgroup$ – sani Jul 1 '16 at 18:03
  • $\begingroup$ I think it is the answer of this -why there is no derivative at sharp point of a curve? also @Ted Shifrin $\endgroup$ – sani Jul 4 '16 at 16:44
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This really just comes down to how you define the line tangent to a curve at a point. Your implied definition that it is a line which "touches" the curve at the point in question but not any of its neighbors in some small interval just brings up the question, "What does 'touches' mean in this context?" If you just mean "intersect", then your definition does not fit with modern definitions of the concept. Which is fine, but if that's the case, then you're right that there are many tangent lines to most curves at most points. And in fact a curve doesn't even need to be differentiable (or even continuous) to have a tangent line at a point. So it's probably not a very useful definition.

There are generally two different ways of defining the tangent line in modern mathematics:

  1. We just define it as the unique line which passes through the point on the curve and whose slope is the derivative of the curve at the point. Note that the curve has to be differentiable at that point if its derivative exists.
  2. We define the tangent line as the best linear approximation to the function at the point on the curve. See professor Shifrin's answer or my answer to this question for an explanation based on this definition.

Note that both of these definitions are equivalent. Both require the curve to be differentiable at the point, will always identify a unique line (when it exists) tangent to the curve, and will always agree on which line that is.

Now, I did say that there are generally two ways of defining the tangent line, but there are some less often used geometrical definitions as well. Michael Hardy gives a good one in his answer to this question.

And thus the reason we would give as to why there's only one tangent line to a differentiable curve at a given point depends on how you define the tangent line to the curve.

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  • $\begingroup$ Can you please explain how Those two definitions you mentioned are equivalent @Bye_World $\endgroup$ – sani Jul 10 '16 at 3:11
  • $\begingroup$ Because as it turns out the best linear approximation is the line which passes through the point on the curve and whose slope is the derivative of the curve at the point. Click the link entitled "my answer to this question" to see a derivation. $\endgroup$ – user137731 Jul 10 '16 at 3:13
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But you have to consider the tangent of a singular point. The tangent only touches a specific singular point and does not touch the neighbour points. There is only a unique tangent which meets this condition. The picture below shows that only one straight line can touch the red point without touching one of the blue points. Any other straight line would touch or intersect one of the neighbour points. With some effort one can imagine this situation for infinitesimally small points.

enter image description here

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    $\begingroup$ What do you mean ? Please be more specific. $\endgroup$ – callculus Jul 1 '16 at 16:51
  • $\begingroup$ Read my question.. $\endgroup$ – sani Jul 1 '16 at 16:52
  • $\begingroup$ Read my answer. You have to consider one single point. Any other case doesn´t make sense. And don´t be so rude. $\endgroup$ – callculus Jul 1 '16 at 16:56
  • $\begingroup$ @callculus What about a line perpendicular to your given one? It wouldn't touch any point on the curve (in this region) except the red one. Really OP's (and your) idea of the tangent line as the unique line which only touches the curve once is full of problems. See lulu's comment on the question. $\endgroup$ – user137731 Jul 1 '16 at 16:56
  • $\begingroup$ @Bye_World It will intersect one of the points. $\endgroup$ – callculus Jul 1 '16 at 16:58
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The value of the derivative is unique (if it exists) because it is defined as a limit, and the value of a limit is unique. Indeed, if we assume that $c$ and $b$ are distinct and

$$\lim_{x\to a}f(x) = c, \lim_{x\to a}f(x) = b$$ Then we can pick $\epsilon = \frac{|b-c|}{2}$, and so there exists $\delta$ so that $|x-a|<\delta$ implies both $|f(x) - b| < \epsilon$ and $|f(x) - c| < \epsilon$. But then $$|b-c| = |b-f(x)+f(x)-c| \leq|b-f(x)|+|c-f(x)|<2\epsilon=|b-c|$$ Which is a contradiction. So, the value of a limit (and therefore the value of a derivative, if it exists) is unique.

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  • $\begingroup$ Its not my answer.Read my qus carefully. $\endgroup$ – sani Jul 1 '16 at 16:23
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As @Ted Shifrin pointed, you define the derivative as the best linear approximation for the function in a point. In particular, when you refer to a real-valued function of real variable, which I assume is your concern from the description, you can speak about its slope.

That slope is the result of a limit, which, because of the properties of the topology in $\mathbb{R}$, must be unique. That means that if you assume there are two tangent lines for a curve in a single point then its derivative in the point cannot be defined, since the value of the limit will depend on how you approach the point.

Tangent lines are only (roughly speaking) equivalent to differentiation in differentiable functions.

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  • $\begingroup$ I do not agree. When differentiation exists at any point. This means there exists a tangent at that point whose slope is equal to the differentiation.This is not saying that there will not exist any other tangent. There may be other tangent at that point whose slope is different from the differentiation value $\endgroup$ – sani Jul 1 '16 at 16:30
  • $\begingroup$ Well than I'm afraid I must disagree with you on that - and we've turned a mathematics discussion into a philosophic one. If you define tangent as you did, that's the only answer you will ever get. Maybe you should be more explicit on what you understand by tangent. $\endgroup$ – cronos2 Jul 1 '16 at 16:41
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Imagine you have a graph of a function, you can zoom in at the point in question further and further. For differentiable function, as your zoom factor grows to infinity, sooner or later the graph starts to appear more and more straight-like. The slope of this (almost) straight line is the value of derivative.

That means if you had a tangent line (tangent according to a common definition) at this point then eventually, as the zoom factor grows, you couldn't tell the two lines apart. This is a graphical way to say that derivative is the best linear approximation for the function at a given point (as it was said in another answer).

Now let's try to find another "tangent". Let's plot an "allegedly tangent" line with another slope through the point. As we zoom in, the graph of the function resembles the real tangent more and more. The "allegedly tangent" goes with different slope, so the two lines must intersect. The intersection angle may be small but it's not zero.

You may define "touch", redefine "tangent" and so on. My point is: for differentiable function there is exactly one straight line you get while zooming in infinitely at a given point of the graph. As there is only one, it is somewhat special, therefore mathematicians have a specific term for it: the tangent at this point.

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