9

So far I assumed the derivative of a function with cases such as this one:

$$f(x) = \begin{cases}x^2\sin x^{-1} & \text{ if } x \neq 0\\ 0 & \text{ else }\end{cases}$$ would be the cases of the derivatives.

So, for $f'$ I would get:

$$f'(x) = \begin{cases} 2x \sin{\left( \frac{1}{x} \right)} -\cos{\left( \frac{1}{x} \right)} & \text{if } x \neq 0\\ 0 & \text{else} \end{cases}$$

And for $f''$

$$f''(x) = \begin{cases} \frac{(2x^2 - 1)\sin{\left( \frac{1}{x} \right)} - 2x\cos{\left( \frac{1}{x} \right)}}{x^2} & \text{if } x \neq 0\\ 0 & \text{else} \end{cases}$$

However, supposedly $f$ is only differentiable once and not twice. Therefore I must have made a mistake here, but I am at a loss as to what that would be. Is the derivative of cases not the cases of the derivatives?

Martin Sleziak
  • 53,687
user3578468
  • 1,371

2 Answers2

8

"The derivative of the cases is the cases of the derivatives", as you write, on any point internal to an interval on which a single case applies. But at any point where two intervals relative to two different cases meet the derivative might not even exist, even if it exists at any point arbitrarily near to it (and even if it approaches the same limit from the right and the left, in fact). Think about $g(x)$ defined as $1/x$ for $x\neq 0$ and as $0$ for $x=0$: is there a derivative in $0$?

To make this all more formal, check the definition of derivative as a limit, and note that in general you really need "a little space around" a point in which only a single case applies to find it.

Anonymous
  • 5,789
5

How you concluded $f'(0)=0$ you made no attempt to explain. Remember that $$ f'(0) = \lim_{h\to0} \frac{ f(0+h) - f(0) } h. $$ You'll probably need to squeeze in order to find the limit. Next you have $$ f''(0) = \lim_{h\to0} \frac{f'(0+h) - f'(0)} h. $$ And again you'll probably have to squeeze. However, it's not hard to see without doing that, that $f'$ is not continuous at $0$ since it approaches no limit at $0$ because of the way it oscillates. Therefore it cannot be differentiable at $0.$

Michael Hardy
  • 1
  • I concluded $f'(0) = 0$ because when $f(x) = c$ then $f'(x) = 0$. How is that incorrect? – user3578468 Jan 14 '17 at 03:15
  • To see why this argument is incorrect, consider the function $g(x)=1/x$ for $x\neq 0$, and $g(x)=0$ for $x=0$. Can you conclude that the derivative in $0$, $g'(0)$, exists and is $0$ because $g(0)$ is "constant" on the point "0"? – Anonymous Jan 14 '17 at 03:28
  • no, hm. Now I see what you mean with "internal to an interval". – user3578468 Jan 14 '17 at 03:33
  • The statement that when $f(x)=c$ then $f'(x)=0$ has to be understood for what it says, not just obeyed unthinkingly. It means $f(x)=c$ for all values of $x$ in some interval. Not just at an isolated point. – Michael Hardy Jan 14 '17 at 19:21
  • @user3578468 : . . . . . or to put it another way, suppose $f(x) = x^3.$ Then $f(2) = 8.$ Since $8$ is a constant, one would conclude that $f'(2) =0.$ But in fact $f'(2) = 3\cdot 2^2 = 12.$ "Constant" must be taken to mean constant in some interval, not "constant" at one point. – Michael Hardy Nov 14 '17 at 01:12