I have to derive such a simple function $y=\sqrt{-x}$, which can be expressed also like this - $y=\sqrt{-1}\sqrt{x}$. Since these functions mean the same I assume their derivatives must be equal too, but they aren't. Why is it so?
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$\sqrt{-x}$ is not the same thing as $\sqrt{-1}\times \sqrt{x}$ in the case that $x$ is negative. – JMoravitz Mar 19 '18 at 22:15
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x is real or complex? – user Mar 19 '18 at 22:15
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x is a real number – Martins2018 Mar 19 '18 at 22:16
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Are you certain that they aren't the same? Expressions may look quite different and still actually be the same. Compare $2+2$ and $4$, for instance. They look completely different, but actually represent the same number. Maybe it's the same with your two derivatives. We can't tell because you haven't shown us what you got. – Arthur Mar 19 '18 at 22:17
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Related: Why $\sqrt{-1\times -1}\neq \sqrt{-1}^2$ – JMoravitz Mar 19 '18 at 22:20
4 Answers
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Remember, the property:
$$\sqrt {ab} = \sqrt a \sqrt b$$
Only applies if $a, b \geq 0$. Thus, the conclusion that:
$$\sqrt{-x} = \sqrt {-1} \sqrt x$$
Is incorrect because $-1 \not\geq 0$.
A common false proof that stems from this problem is as follows:
$$\begin{align} 1 &= \sqrt 1 \\ &= \color{red}{\sqrt {-1 \cdot -1}} \\ &= \color{red}{\sqrt{-1} \cdot \sqrt{-1}} \\ &= i^2\\ &= -1\end{align}$$
Andrew Li
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Just take the derivative normally, remember $(-x)^{1/2}$ so taking the derivative is just $$\frac12(-x)^{-1/2}(-1)=-\frac{1}{2\sqrt{-x}}$$
Teh Rod
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The problem is the domain of the function.
The first thing that comes to mind since you use the variable $x$ is that $x$ is a real number (as opposed to the complex $z$). Hence $\sqrt{-x}$ is a real-valued function of a real variable defined for $x\leq 0$.
Now when you write $\sqrt{-1}\sqrt{x}$, this is still a function of one real variable, but it is now complex-valued, and the domain is now $\{x\geq 0\}$!
So, it is not the same function (at all) and there is no wonder the derivative is not the same.
Morality: be careful about domains when you define a function.
Arnaud Mortier
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The identity $\sqrt{ab} = \sqrt{a}\sqrt{b}$ is true only for positive numbers. There's a "proof" 1 = -1 that goes as follows:
$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{(-1)} \sqrt{(-1)} = i*i = -1$
The problem is that $\sqrt{x}$ denotes a number y such that $y^2 = x$. But there are two such numbers. For instance, if x = 4, then $y^2 = x$ has the solution y = 2 and y = -2. So $\sqrt{x}$ is taken to mean the positive number such that this is true, which is known as the principle square root. But when x is negative, there is no positive solution; both solutions are imaginary. So the principle square root is labeled "i", but it's arbitrary which solution is taken; there's no way to distinguish i and -i (a full discussion of this gets into Galois groups). So when we say "The square root of -1 is i", what we mean is "There are two numbers that square to -1, and we've labeled one of them i". Because of this arbitrary choice of the value of $\sqrt{(-1)}$, the properties that square root has over positive numbers break down.
Acccumulation
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