# If function $f$ is differentiable at a point $a$, does it imply that $f'(a)$ exists?

If function $f$ is differentiable at a point $a$, does it imply that $f'(a)$ exists?

I want to proof that if function $f$ is differentiable at a point $a$, then the function is continious at this point. I want to do this by using the multiplication rule for limits. For that I need to show that function $f$ has a finite limit at point $a$. I have read that function if is differentiable at a point $a$ if $f'(a)$ exists. However, it seems that it is not exactly what I need.

Thank you for your asnwers. I was a bit confused by how the definiton phrased.

I have an additional question: how does one know when "if" means "if and only if" and "if" means "if"?

• $f'(a)$ is, by definition, the derivative of $f$ at $a$. So yes – TomGrubb Dec 31 '15 at 17:02
• Downvote. This is a most trivial notational question, the answer to which can be found practically anywhere. – MathematicsStudent1122 Dec 31 '15 at 17:05
• I disagree with the down-vote and the vote to close the question. How to prove that if $f$ is differentiable at $a$ then $f$ is continuous at $a$ is a reasonable question. And it is unreasonable to expect students posting things like this never to be confused about some of the details. ${}\qquad{}$ – Michael Hardy Dec 31 '15 at 17:14
• "If" never means "if and only if", however, "if and only if" is sometimes written as "iff". – Reinhild Van Rosenú Dec 31 '15 at 21:57

The definition of "$f$ is differentiable at a point $a$" is that "$f'(a)$ exists". So $f$ is differentiable at $a$ if and only if $f'(a)$ exists.

(When you read that "function $f$ is differentiable at a point $a$ if $f'(a)$ exists", that was probably using the standard convention in mathematics that "if" actually means "if and only if" when making a definition. That is, if you write "I define $f$ to be differentiable at $a$ if $f'(a)$ exists", you actually mean that $f$ is differentiable at $a$ if and only if $f'(a)$ exists.)

• Thank you for your answer. Then how does one know when "if" means "if and only if" and "if" means "if"? – user048983486 Dec 31 '15 at 17:12
• It means "if and only if" when (and only when) it appears at the start of a definition, as in the example I gave. The rationale is that if it actually just meant "if", you wouldn't be providing a complete definition at all--you would just be mentioning some sufficient conditions for your newly defined property to hold. – Eric Wofsey Dec 31 '15 at 17:15
• Why not to use "if and only if"? It seems to be less confusing. – user048983486 Dec 31 '15 at 17:19
• It's just a standard convention; people use it because it is what they are used to hearing and reading, and it is briefer. It is unfortunate how it can cause confusion among beginners, but it is not so easy to change the habits of thousands of mathematicians who have been trained to follow the convention. – Eric Wofsey Dec 31 '15 at 17:21
• In the definition of the limit we have the following part: $0<|x−c|<δ$ implies $|f(x)−L|<ϵ$. Am I correct to assume that the implication actually means implication? – user048983486 Dec 31 '15 at 17:32

To say $f$ is differentiable at $a$ means that $f'(a) = \lim\limits_{h\to0} \dfrac{f(a+h) - f(a)} h$ exists. That is a definition, not something requiring proof.

If you want to prove that if $f$ is differentiable at $a$ then $f$ is continuous at $a$, you can do so as follows.

First, let us note that trivially $\lim\limits_{h\to0} \dfrac{f(a+h) - f(a)} h$ is the same as $\lim\limits_{x\to a} \dfrac{f(x)-f(a)}{x-a}$. That that limit exists is another way of saying what it means to say $f$ is differentiable at $a$.

Since $\lim\limits_{x\to a} \dfrac{f(x)-f(a)}{x-a} \vphantom{\dfrac1{\displaystyle\sum}}$ exists (and that means it's a finite number, not $+\infty$ or $-\infty$) and $\lim\limits_{x\to a} (x-a)$ exists, the limit of the product exists and is equal to the product of the two limits: $$\lim_{x\to a} \left(\frac{f(x)-f(a)}{x-a} \cdot (x-a)\right) = \left(\lim_{x\to a} \frac{f(x)-f(a)}{x-a}\right) \cdot \left(\lim_{x\to a} (x-a) \right).$$ The first limit is some finite number (since $f$ is differentiable at $a$) and the second is $0$. Therefore $$\lim_{x\to a} (f(x)-f(a)) = 0.$$ And given that, one can show that $$\lim_{x\to a} f(x) = f(a),$$ so $f$ is continuous at $a$.

• Should your displaymath read $\lim_{x\to a} \cdots$ ? – Strants Dec 31 '15 at 17:27

Yes, it implies that $f'(a)$ exists and $f(x)$ is continuous at $x=a$. That's the definition of differentiability.

• I think this is more suited as a comment than an answer – Learnmore Dec 31 '15 at 17:03
• Why? It answers the question, doesn't it? – fleablood Dec 31 '15 at 17:05
• @learnmore: I think your observation is more suited as an unspoken thought than a comment. – TonyK Dec 31 '15 at 17:06
• @kamil09875 Thank you for your answer. "More generally, if x0 is a point in the domain of a function f, then f is said to be differentiable at x0 if the derivative f ′(x0) exists". This is the definition from wikipedia. Why does it say "if the derivative f ′(x0) exists" and not "if and only if the derivative f ′(x0) exists"? – user048983486 Dec 31 '15 at 17:08
• @user048983486: I have addressed that in my answer. Saying "if" instead of "if and only if" is just a convention when stating definitions. – Eric Wofsey Dec 31 '15 at 17:10