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As part of a high school math problem, I have to figure out if the function $y=x\sqrt{x(4-x)}$ has a derivative when $x=0$. My working out seems to indicate this derivative exists and is zero. Yet, when I want to check on WolframAlpha, I get this : derivative undefined... https://i.stack.imgur.com/t4iB5.jpg Can you explain ?

Plot here : https://i.stack.imgur.com/8MRII.jpg

imranfat
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shadok
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    @Shadok. It would be useful if you share your computations with us... – imranfat Oct 21 '16 at 17:14
  • @MichaelLee What? Of course the derivative at 0 (from the right) exists. – Did Oct 21 '16 at 17:14
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    @Did Do the left and right derivatives of $\lvert x \rvert$ exist at $x = 0$? If they don't agree, then the derivative of $\lvert x \rvert$ does not exist at $x = 0$. Similarly, the left and right derivatives of $x\sqrt{x(4-x)}$ do not agree if we are constrained to the reals. – Michael L. Oct 21 '16 at 17:20
  • @MichaelLee : As a real-valued function of a real variable, this function has domain $[0,4]$. The derivative exists at $x=0$ and is zero. It doesn't matter that the domain happens to have an endpoint there. – MPW Oct 21 '16 at 17:24
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    @MPW This is only true if you choose a definition of differentiable such that a function is called differentiable on an endpoint of its domain if it has a one-sided derivative there. Certain authors (e.g. Rudin) choose a definition like this, but it's not universal. As before, if you call $[0, \infty)$ the domain of $\lvert x \rvert$, then $\lvert x \rvert$ is "differentiable" at $x = 0$ under this definition. – Michael L. Oct 21 '16 at 17:32
  • @MichaelLee : Of course $|x|$ is differentiable at $x=0$ if you restrict the domain to $[0,\infty)$. It is indistinguishable from $g(x)=x$ on the same domain. The thing that makes $|x|$ non-differentiable at $x=0$ is precisely the difference in behavior on either side of $0$. Restricted, this doesn't happen. – MPW Oct 21 '16 at 17:47
  • @MichaelLee : The definition of $f'(x)$ is $\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$. "$y\to x$" means $y$ approaches $x$ through values in the domain. If it happens that all of the domain is on one side of $x$, so be it. The derivative still either exists or doesn't, precisely when this limit exists or doesn't. For the limit to exist, it must have the same value for any such approach $y\to x$ through the domain. – MPW Oct 21 '16 at 17:55
  • @MPW I'll say it again... your definition is not universal. See the second half of page 3: https://math.dartmouth.edu/opencalc2/cole/lecture21.pdf. In fact, several authors even reject the notion of a function being "differentiable on a closed interval." – Michael L. Oct 21 '16 at 19:20
  • @MPW Here's another example, in which it is stated that the "usual convention" is to leave the derivative at the endpoint of a function's domain undefined: https://books.google.com/books?id=0XX9eI1SneoC&pg=PA245&lpg=PA245. – Michael L. Oct 21 '16 at 19:28
  • @MichaelLee Sorry but how does your first comment supposed to apply to the function at hand? Note that, for example, $g(x)=x\sqrt{|x|(4-x)}$ is differentiable at $0$ hence the problem is definitely not "the denominator" or comparing "the left and right derivative" or a "discontinuity" but that $f$ is defined on $[0,\infty)$ instead of on a (bilateral) neighborhood of $0$. And once again $f$ is differentiable from the right at $0$. Perhaps you misread $f(x)$ for $\sqrt{x(4-x)}$? Anyway, as it is, your first comment is rather misleading, as are your post hoc rationalizations invoking $|x|$. – Did Oct 21 '16 at 21:07
  • @MichaelLee Did you downvote Nicolas' answer by any chance? – Did Oct 21 '16 at 21:08
  • @Did I did not. – Michael L. Oct 21 '16 at 22:58
  • @Did You're right, my first comment was unfounded. However, my point still stands that the function is only differentiable if we either (1) consider $y$ from the reals to the complexes, or (2) take the derivative at the endpoint of the domain to be defined as its one-sided limit. I've certainly seen both (either defined as above, or left undefined entirely), and we have to be specific as to which definition we use. I generally prefer the latter (I don't like my derivatives taking different values when considered on submanifolds of their domains than when considered on the whole domain). – Michael L. Oct 21 '16 at 23:11
  • @Did I've deleted the first comment so that it wouldn't continue to be misleading. – Michael L. Oct 21 '16 at 23:22

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Seems like WA uses the standard derivation formulas, tries to substitute $0$ to $x$, and finds a division by $0$.

The theorem about $\sqrt u$ tells you that this function is differentiable when $u$ is differentiable, and strictly positive. If you apply it to your function, you'll find that the theorem applies only on $]0,4[$ (and in fact, the function has no derivative at $4$).

So if you want to check wether you can derive at $0$, you have either to

1) get back to the definition : $$\frac{f(x)-f(0)}{x-0}=\sqrt{x(4-x)}\xrightarrow[x\to 0]{}0$$ 2) use the limit of the derivative (I don't know the name of the theorem in english) : if $f$ is continuous on $[a,b]$, differentiable on $]a,b]$ and if $f'(x)$ has a finite limit $\ell$ when $x$ tends to $a$, then $f$ is differentiable at $a$ and $f'(a)=\ell$. This theorem applies here too.

Nicolas FRANCOIS
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