Doubt on how to express $\sqrt{x}^2$ on derivative function... which is the right answer?

Question

I was requested to find the derivative of $f(x)=\frac{x}{\sqrt{1-x^2}}$. When I was working on this, I found an expression of the form $\sqrt{1-x^2}^2$, which I translate to $|1-x^2|$. My calculus teacher is very insistent on the fact that $\sqrt{x}^2=|x|$, and I understand the reason of this result. Nevertheless, when I found the derivative, I used Symbolab to check if my answer was right, and noticed that it ignored the absolute value, and simply stated that $\sqrt{1-x^2}^2=1-x^2$. I can't find a mathematical reason for this result, which of course leads to a different derivative function. Is Symbolab wrong, or am I? This is how I found the derivative.

$f'(x)=\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{\sqrt{1-x^2}^2}$.

I skipped the previous steps because aren't relevant; up to this point, Symbolab and I agree. But I take the following course of action: let $u=1-x^2$.

$f'(x)=\frac{\sqrt{u}+\frac{x^2}{\sqrt{u}}}{|u|}=\frac{|u|+x^2}{\sqrt{u}|u|}$.

Symbolab, on the other hand, by getting simply $u$ each time I got $|u|$, gets

$f'(x)=\frac{1}{\sqrt{u}*u}$

Which is the right answer?

Answer 1

Your function is only defined when $1-x^2 >0$. Therefore, on the domain of $f(x)$, it is always the case that $\sqrt{1-x^2}^2 = 1-x^2$. Also notice that the result of Symbolab is only defined when $1-x^2>0$, so your derivative and Symbolab's derivative are exactly the same.

Answer 2

There's notational ambiguity here. What do you mean by $$\sqrt x ^2,$$ that is?

If it means $$\sqrt{x^2},$$ then your teacher is right to insist that it is equal only to $|x|$ provided we know nothing else about $x.$ This follows since that expression is defined for all $x,$ and whatever it evaluates to must not be negative, by definition.

On the other hand if what you mean is $$(\sqrt x)^2,$$ then this always means $x$ since $x$ must not be negative in the first place for the expression to represent a real number. I don't think your teacher would insist that this latter be $|x|.$

Thus, you need to be clear what you mean by your notation.

Answer 3

Neither

$$|1-x^2|$$

nor

$$1-x^2$$

are fully equivalent to

$$\left(\sqrt{1-x^2}\right)^2, $$

just like neither

$$|x|$$ nor $$x$$ are fully equivalent to $$\left(\sqrt x\right)^2.$$

Because the domains are different.

Answer 4

Your teacher is wrong to insist that $\sqrt x^2=|x|$. If the square root symbol is to have any meaning at all, its square must be equal to the thing inside, i.e., $\sqrt x^2=x$. If $x$ is positive, there is no problem, of course, since $|x|=x$ when $x\ge0$. If $x$ is negative, you have to decide whether to allow imaginary numbers or simply declare the square root does not exist. If you go the former route, it's standard to write $\sqrt{-1}=i$, with $i^2=-1$, so that we have $\sqrt{-1}^2=-1$. If you go the latter route, the expression $\sqrt x^2$ does not make sense, since you are trying to square something that does not exist! Whichever way you go, $\sqrt x^2\not=|x|$ when $x$ is negative.

When doing calculus with expressions involving the square root symbol, it's best to be careful about the domain of the function, and avoid square roots of negative numbers, since calculus, at least as it's organized and presented in a first course, is limited to real numbers. A lot of the theorems extend to complex numbers, but the theory needs to be carefully redeveloped to be sure of the things that are still true (and see which things aren't). The effort is ultimately worthwhile, since the calculus of complex variables is stunningly powerful and wonderfully beautiful.

In sum, if your teacher insists that $\sqrt x^2=|x|$ for all real numbers $x$, ask them if $i^2=1$. As Allawonder has already noted, it's possible your teacher really has the formula $\sqrt{x^2}=|x|$ in mind, which is true for all real numbers $x$.