The Concept

Definition 4.1.1  
A function $f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto f(x)$ is called differentiable at a point $x$ if the limit

$$f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$ (4.1.1)

exists. The function $f'$ is then called the derivative of $f$ in $x$. Often $f'(x)$ is written as $\frac{df}{dx}(x)$

Remark 4.1.2  
The fraction

$$\frac{f(x+h)-f(x)}{h}$$ (4.1.2)

is the slope of a line going through the points $(x,f(x))$ and $(x+h,f(x+h))$. Note that both points are on the graph of the function $f$. Letting $h$ approach zero then boils down to calculating the slope of the tangent line in $(x,f(x))$.

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This geometric interpretation will often be used when trying to understand a specific example.

Theorem 4.1.3  
A function that is differentiable at a certain point $x$ is also continuous at that point $x$.

Proof.
First note that for any $h \not= 0$, we have

$$f(x+h) = f(x) + \frac{f(x+h) - f(x)}{h} \cdot h$$ (4.1.3)

Nothing mysterious happened here, it is all fairly elementary algebra. Taking the limit on the left and right hand side then gives

$$\lim_{h \to 0}f(x+h) = \lim_{h \to 0}\left[ f(x) + \frac{f(x+h) - f(x)}{h} \cdot h \right]$$ (4.1.4)

which reduces to

$$\lim_{h \to 0}f(x+h) = f(x) + f'(x) \cdot \lim_{h \to 0}h$$ (4.1.5)

Therefore $$\lim_{h \to 0}f(x+h) = f(x)$$ which by definition means that $f$ is continuous in x. $\qedsymbol$

Remark 4.1.4  
The previous theorem states that differentiable functions are continuous, but then the natural question to ask is if the converse is true. Quite a lot of functions that are continuous over their domain are differentiable over their domain, in fact all the functions in the list of continuous functions (3.1.4) are differentiable over their domain. In general, however, this will turn out not to be the case.
Exactly as with the concept of continuity, there are two reasons why a function can be non-differentiable in a certain point: either the limit needed to calculate the derivative goes to infinity, or it is not defined (often because taking the limit from the left gives a different result then taking the limit from the right).

• The limit is infinite: it is very easy to come up with discontinuous functions which are then not differentiable at the point of discontinuity (e.g $\tan(x)$ or $\frac{1}{x^n}$ for any $n$). A continuous function, can however also have this glitch; take e.g. the function

  $$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto \sqrt{\vert x\vert}$$ (4.1.6)

  
  
  The graph of this function hints quite convincingly that it is a continuous function, but it is also clear that the tangent to the graph in 0 is infinite, so the function is not differentiable in 0.
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• The limit is undefined: Take e.g the function

  $$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto \vert x\vert$$ (4.1.7)

  
  
  As the picture below indicates, this function is clearly continuous, but it is also easy to see that the left limit in 0 is $-1$ while the right limit in 0 is $1$.
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  It is not always obvious that the left and right limit are different from one another. Take a look at the function

  $$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: \... ...\frac{1}{x}) & \text{if $x\not=0$} \\ x \mapsto 0 & \text{if $x=0$} \end{cases}$$ (4.1.8)

  
  
  The behaviour of this function around 0 looks like:
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  To the untrained eye, the picture suggests a very moderate behaviour around zero, but alas, this is not really the case. On one hand we have

  $$0 \leq \left\vert \limx0 x \cdot \sin\left(\frac{1}{x}\right) \right\vert \leq \limx0 \vert x\vert = 0$$ (4.1.9)

  
  
  so $$\lim_{x \to 0} f(x) = 0 = f(0)$$ and therefore $f$ is continuous in 0. But on the other hand,

  $$\lim_{h \to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h \to 0}\frac{h\sin\left(\frac{1}{h}\right)-0}{h} = \lim_{h \to 0}\sin\left(\frac{1}{h}\right)$$ (4.1.10)

  
  
  which is not defined. So $f$ is not differentiable in 0.