# Derivative of irrational function is undefined at zero, but it seems weird for me.

There is an example that confuses me:

$$f(x) = x \sqrt x$$

Taking the derivative:

$$f'(x) = (x)' \sqrt x + (\sqrt x)'x = \sqrt x + \frac {x}{2\sqrt x} = \frac {2x + x}{2 \sqrt x} = \frac{3x}{2 \sqrt x}$$ So far it makes sense. The derivative is undefined at zero because we can't put 0 in denominator. But on the other hand, if we use the power rule:

$$f'(x) = (x \sqrt x)' = (x^{1}x^{0.5})' = (x^{1.5})' = 1.5x^{0.5} = \frac{3 \sqrt x}{2}$$ Which IS actually defined at zero.

Then I got it: $$\frac{3x}{2 \sqrt x} = \frac{3x}{2 \sqrt x} \frac{\sqrt x}{\sqrt x} = \frac{3x \sqrt x}{2x}$$ But x could possibly be 0, so we can't divide by it: $$\frac{3x \sqrt x}{2x} \not= \frac{3 \sqrt x}{2}$$

Why it appears when I use the power rule? What am I missing?

• Your function $f$ is a priori not defined for $x < 0$. The maximal domain is $[0, \infty)$, so it is not differentiable at 0. (For differentiability you need to be able to approach from both sides). So in fact, $x \neq 0$ and you can divide by it so both solutions should agree. Feb 9, 2023 at 6:47
• Your function $f,$ defined indeed on $[0, \infty),$ is (right-)differentiable at $0$ and $f'(0)=0,$ by the power rule, which is more efficient here than your first method. Btw, would you be able to calculate $f'(0)$ directly from its definition as a limit? (as $x\to0^+,$ of $(f(x)-f(0))/(x-0)$) Feb 9, 2023 at 6:59
• @noam.szyfer But change $x\sqrt x$ to $x\sqrt[3]x$ and the OP's issue remains; so I don't think this is really the crux of the question. Feb 9, 2023 at 7:37

The power rule (or the definition) correctly calculates that $$f'(0)=0$$. So the question is, why don't we get this answer using the product rule?

If we look closely at the product rule, this is what it states: if both $$g(x)$$ and $$h(x)$$ are differentiable at $$x=a$$, then $$g(x)h(x)$$ is also differentiable at $$x=a$$ and the derivative equals $$g'(a)h(a)+g(a)h'(a)$$.

This is an if-then statement, not an if-and-only-if statement. It doesn't make any claims about what happens if one or both factors is not differentiable at that point.

Since $$\sqrt x$$ is not differentiable (even right-differentiable) at $$x=0$$, the hypotheses of the product rule are not satisfied, and so carrying out the computation gives no information about the actual derivative of $$x\sqrt x$$.

[In general we should always remember: formulas have hypotheses!]

• OK, thank you, I get it. But what should you do if you see something like $(x+2) \sqrt {x+3}$? The power rule doesn't work here. Feb 12, 2023 at 7:01
• I assume you mean the derivative at $x=-3$, for which the product rule is similarly inconclusive. I suggest going back to the actual definition of the derivative, which will show that this function is not right-differentiable at $x=-3$. Feb 12, 2023 at 7:56
• I mean what the derivative formula of this looks like. I also asked my teacher (she's a professional mathematician) this question (about $(x \sqrt x)'$). She says that you can't differentiate $x \sqrt x$ at $x = 0$, it's undefined at that point, it comes from the definition of derivative. You can't approach the limit from the left. Feb 13, 2023 at 18:42

As Anne Bauval pointed out, your function is right-differentiable at $$\,x=0\,$$ and we can calculate $$\;f’_+(0)\;$$ directly from its definition as a limit, that is,

$$f’_+(0)=\lim\limits_{x\to0^+}\dfrac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0^+}\dfrac{x\sqrt x}x=\lim\limits_{x\to0^+}\sqrt x=0\;\;,$$

hence , $$\;\;f’_+(0)=0\,.$$

Nevertheless, you can get this result from your first calculation of the derivative

$$f’(x)=\dfrac{3x}{2\sqrt x}$$

by using the following theorem :

If a function $$\;f:[a,b]\to\Bbb R\;$$ is continuous on $$[a,b]$$, differentiable on $$(a,b)$$ and there exists $$\lim\limits_{x\to a^+}f’(x)=l\in\Bbb R\;,\;$$ then the function $$f$$ is right-differentiable at $$\,x=a\,$$ and $$\,f’_+(a)=l\,.$$

Therefore, since $$\;\lim\limits_{x\to0^+}\dfrac{3x}{2\sqrt x}=\lim\limits_{x\to0^+}\dfrac{3\sqrt x}2=0\;,\;$$ it results that your function is right-differentiable at $$\,x=0\,$$ and $$\;f’_+(0)=0\,.$$

It is very important to understand that if an expression of a derivative is not defined in a point $$x_0$$, it does not necessarily imply that the function is not differentiable at that point $$x_0$$, indeed it could be possible to find out that actually the function is differentiable at that point $$x_0$$ by applying another method or by using the theorem I wrote in this answer.