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?
 A: 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.
A: 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!]
