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What does it mean to take a derivative of a point-wisely defined function, in this case:

$\ f(x) = \begin{cases} \frac{\sin x}{x} & \text{if } x ≠ 0 \\ 0 & \text{if }x=0 \end{cases} $

mavavilj
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2 Answers2

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For each $x \neq 0$, $f'(x)$ is just the derivative of $\sin{(x)}/x$, since $$\lim \limits_{h \to 0} \dfrac{f(x + h) - f(x)}{h} = \lim \limits_{h \to 0} \dfrac{\dfrac{\sin{(x + h)}}{x + h} - \dfrac{\sin{(x)}}{x}}{h}.$$

But, you have to worry about if the derivative exists at $0$. To be differentiable, it has to be continuous first, so you should check that $\lim \limits_{x \to 0} f(x) = f(0)$, i.e., $\lim \limits_{x \to 0} \dfrac{\sin{(x)}}{x} = 0$.

(Also, recall that $\lim \limits_{x \to 0} \dfrac{\sin{(x)}}{x} = 1$ which you can verify using L'Hopital's rule).

layman
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  • Is that not a contradiction? $\lim \limits_{x \to 0} \dfrac{\sin{(x)}}{x} = 0 = 1$ – mavavilj Oct 25 '15 at 12:51
  • @mavavilj Well, the fact that the limit is $1$ means it is not $0$. What does that say about the function $f$? What does it imply about the derivative at $0$? – layman Oct 25 '15 at 12:54
  • Its derivative is discontinuous? – mavavilj Oct 25 '15 at 13:15
  • @mavavilj No, the function itself is discontinuous at $x = 0$. If it is discontinuous at $x = 0$, then the derivative cannot exist at $x = 0$, so the derivative is undefined at $x = 0$. – layman Oct 25 '15 at 13:30
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It means (I'm assuming your question is at $x = 0$) applying the definition of the derivative to the function in question. That definition is not "apply all the rules of differentiation", but rather, $$ f'(0) = \lim_{h \to 0} \frac{ f(0 + h) - f(0) }{h}. $$ If you write out that limit in this case, you get $$ f'(0) = \lim_{h \to 0} \frac{ \frac{\sin h}{h} - 0 }{h}. $$

Now you have to actually compute that limit (if it exists).

"But why can't I just use the formulas?" you might ask. Well, look at the hypotheses behind each formula, and you may perhaps discover why.

John Hughes
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