Differentiability and its relation with continuity

Differentiability

The derivative does not always exists, since it can be the case that at a given point the limit fails to exist.

Nevertheless, there exist certain conditions that allow us to assure the existence of the derivative function.

When these conditions are met we say that the function $f (x)$ is differentiable.

Consider the function defined by: $$\displaystyle f(x)=\left\{ \begin{array}{rcl} x &\mbox{ if } & x<0 \\ x^2& \mbox{ if }& x \geq 0 \end{array} \right.$$ Clearly, the point $x=0$ is special. Does the derivative exist in this point?

To see that the derivative fails to exist at $x=0$ try do the following: try to find the value of the derivative approaching to point $x=0$ from the left and then from the right.

If two values exist and coincide we will say that the function is differentiable at $x=0$.

Let's see an example:

First we compute the derivatives from each side:

When we approach $x=0$ from the left we have: $$\displaystyle f'(0^-)= \lim_{\Delta x \ to 0^-}\frac{f(0+\Delta x)-f(0)}{\Delta x}$$

(We do not consider the case where $\Delta x$ tends towards zero from any direction, but rather we must bear in mind that $\Delta x$ is always negative).

When approaching from the left, we are in the case $f (x) =x$, therefore the computation is as follows: $$\displaystyle f'(0^-)=\lim_{\Delta x \to 0^-} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim_{\Delta x \to 0^-} \frac{(0+\Delta x)+0}{\Delta x}=1$$ When the function approaches $0$ from the right we have $f(x)=X^2$ the computation is then: $$\displaystyle f'(0^+)=\lim_{\Delta x \to 0^+}\frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim_{\Delta x \to 0^+}\frac{(0+\Delta x)^2+0}{\Delta x}=\lim_{\Delta x \to 0^+}\Delta x=0$$ (Now $\Delta x$ has to be a positive)

In this case, then, the two side derivatives (from the left and from the right) exist but its value does not coincide. Therefore the function $f (x)$ is not differentiable at the point $x=0$.

Summing up, we have obtained a differentiability condition:

'A function is differentiable at a given point if, and only if, the side derivatives exist and they coincide'.

Also, in a more general way, one can imagine that in the peaks or angular points of the functions, the functions are not differentiable.

In other words, a function is differentiable at a given point if it goes smoothly over this point.

Differentiability and continuity

A differentiable function at a point $a$ is also continued at this point $a$.

(The reciprocal need not to be true, that is, a continuous function at a point $a$ needs not to be differentiable at this point)

We are going to study the continuity and differentiability of the following functions.

$$f(x)=\left\{\begin{array} {rcl} 3 & \mbox{ if } & x < 0 \\ x & \mbox{ if } & x\geq 0\end{array} \right.$$

To study the continuity (3 steps: value at the point, limit from the left, and limit from the right): $$\begin{array}{l} f(0)=0 \\ \lim_{x \to 0^-}f(x)=3 \\ \lim_{x \to 0⁺}f(x)=0 \end{array}$$ The value of the function at $x=0$ is zero, by definition.

Nevertheless, the limits from the right and from the left do not coincide at this point. We say, then, that this function is not continuous in $x=0$.

Since the function is not continuous it cannot be differentiable.

$$f(x)=\left\{ \begin{array} {rcl} 0 & \mbox{ if } & x < 0 \\x & \mbox{ if } & x \geq 0 \end{array} \right.$$ The function is similar to the previous one, although now when we compute the limit from the left the result is also zero.

Therefore the function is continuous at $x=0$.

This does not mean that it is also differentiable at $x=0$. To see whether it is differentiable or not we need to do further computations.

To study the differentiabiltiy (2 steps: value of the derivative approaching from the left and from the right)

$$\displaystyle \begin{array}{l} f'(0^-)=\lim_{\Delta x \to 0^-}\frac{f(0)-f(0)}{\Delta x}= \lim_{\Delta x \to 0^-}\frac{0+0}{\Delta x}=0 \\ f'(0^+)=\lim_{\Delta x \to 0^+}\frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim_{\Delta x \to 0^-}\frac{(0+\Delta x)+0}{\Delta x}=1\end{array}$$

The values do not coincide and that's why it is said that the function $f (x)$ is not differentiable at the point $x=0$, although, as has been seen, it is continuous at this point.