Differentiable function at some point

Question

When checking if function is differentiable at point we get the derivative from left and right Why always we get the derivative in this case as in most books and references using differentiation by definition not differentiation rules? And is it wrong if we get the derivative from left and right using the differentiation rules?

Answer 1

The differentiation rules require that your function is differentiable. You want to prove this. Generally what you are saying works out but it is not correct. You can use it 99% of the times to do your work(I mean to solve problems in sciences except math) but your (math) teachers will count it as wrong. Here is an example of your method working(but still not mathematically correct):

f(x) = \begin{cases} x^2 , & x<0 \\ 0 ,& x>=0 \end{cases}

And here is another example where your method does not work:

f(x) = \begin{cases} x^2, & \text{if $x$ is rational} \\ 0x^2, & \text{if $x$ is not rational} \end{cases}

By using the definition of derivative you will find out that this function is differentiable in point 0. However your way is still a great way to test your answer for yourself

Further explanation for the second example: Let:

g(x) = \begin{cases} 1, & \text{if $x$ is rational} \\ 0, & \text{if $x$ is not rational} \end{cases}

Then f(x)=$x^2g(x)$

So you have:

$\lim\limits_{x \to 0} \frac{f(x)-f(0)}{x-0}$=$\lim\limits_{x \to 0} \frac{x^2g(x)}{x}$=$\lim\limits_{x \to 0} xg(x)$

And since -|x|<=xg(x)<=|x| you can use the squeeze theorem to prove that the limit is equal to 0. (And consequently g'(0)=0)