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I'm trying to find the derivative by definition of the following function:

$f(x)=\sqrt{|x|}\sin(x)$

I know that by definition:

$$ f'(x)=\lim_{h\to0}\frac{\sqrt{|x+h|}\sin(x+h)-\sqrt{|x|}\sin(x)}{h} $$

But if I try to find the derivative at $0$ I get:

$$ f'(0)=\lim_{h\to0}\frac{\sqrt{|h|}\sin(h)}{h}=0 $$

which is not true because the derivative $DNE$ at $0$ because:

$$f'(x)=\begin{cases} \sqrt{x} \cos x+\frac{\sin x}{2\sqrt{x}}, \ \ \ \ \ x>0 \\ \sqrt{-x} \cos x-\frac{\sin x}{2\sqrt{-x}}\ \ \ x<0 \\ \end{cases}$$

So how can it be that the derivative exists only when it is calculated by definition?

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  • $\begingroup$ Your final expression for $f'(x)$ looks as if it has limits of $0$ as $x \to 0$ from above or below. Combine this with the continuity of $f(x)$ $\endgroup$ – Henry Jan 19 '16 at 10:57
  • $\begingroup$ You could make the derivative concise by taking $$\frac{d}{dx}|x|=|x|/x$$. Try evaluating the resultant derivative using a limit $\endgroup$ – Prish Chakraborty Jan 19 '16 at 13:07
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You shouldn't say that the derivative does not exist.

Indeed,

$$\lim_{h\to0}\frac{\sqrt{|h|}\sin(h)}{h}=\lim_{h\to0}\sqrt{|h|}\cdot\lim_{h\to0}\frac{\sin(h)}{h}=0\cdot1.$$

As the limit exists, this is the value of the derivative.

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    $\begingroup$ Just a comment to the OP: the first equality holds because both limits on the right exist. If $\lim g$ and $\lim f$ exist, then $\lim f\cdot g$ exists and is equal to $\lim f\cdot \lim g$ $\endgroup$ – 5xum Jan 19 '16 at 9:18
  • $\begingroup$ Yves, I edited my question. $\endgroup$ – CodeNinja Jan 19 '16 at 9:31
  • $\begingroup$ @CodeNinja: when you modify a question after the fact, you should clearly indicate it, otherwise comments and answers can become nonsense. $\endgroup$ – Yves Daoust Jan 19 '16 at 10:03
  • $\begingroup$ I just clarified my intention. $\endgroup$ – CodeNinja Jan 19 '16 at 10:07
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    $\begingroup$ @CodeNinja: you should clearly indicate the change. $\endgroup$ – Yves Daoust Jan 19 '16 at 10:20
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We have $$f(x)=\begin{cases} \sqrt{x} \sin x, \ \ \ \ \ x\geq0 \\ \sqrt{-x} \sin x\ \ \ x<0 \\ \end{cases}$$ and $$f'(x)=\begin{cases} \sqrt{x} \cos x+\frac{\sin x}{2\sqrt{x}}, \ \ \ \ \ x>0 \\ \sqrt{-x} \cos x-\frac{\sin x}{2\sqrt{-x}}\ \ \ x<0 \\ \end{cases}$$ But $$\frac{\sin x}{2\sqrt{x}}\rightarrow 0, \ \ \ \frac{\sin x}{2\sqrt{-x}}\rightarrow 0$$ for $x\rightarrow 0^+$ and $x\rightarrow 0^-$. Moreover, by definition of derivative: $$\lim_{h\to0}\sqrt{|h|}\cdot\lim_{h\to0}\frac{\sin(h)}{h}=0\cdot1.$$

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  • $\begingroup$ but it does not defined at $x=0$ so the derivative does not exist at that point (because you divide by $\sqrt{x}$) $\endgroup$ – CodeNinja Jan 19 '16 at 9:26
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    $\begingroup$ @CodeNinja Not true. The derivative at $0$ is defined as $\lim_{h\to 0} \frac{f(0+h) - f(0)}{h}$. That is the definition. The fact that $\sqrt{x}\cos x + \frac{\sin x}{2\sqrt x}$ is not defined for $x=0$ does not mean that the derivative does not exist. $\endgroup$ – 5xum Jan 19 '16 at 9:30
  • $\begingroup$ I agree with @5xum. Furthermore, the two limits are equal. $\endgroup$ – Mark Jan 19 '16 at 9:32
  • $\begingroup$ @Mark, So I can I know when I can calculate the derivative by a formula (like the chain rule of something else) or that it can be calculated only by definition. $\endgroup$ – CodeNinja Jan 19 '16 at 9:37
  • $\begingroup$ @CodeNinja, if $\exists$ $\lim_{x\to x_0} f'(x)$ then $f'(x_0)=\lim_{x\to x_0} f'(x)$. But beware: this is only a sufficient condition. $\endgroup$ – Mark Jan 19 '16 at 9:55

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