# Derivative of $\sqrt {xy}$?

$\sqrt{|xy|} = 1$

Attempting to find the derivative gives me $$\frac12(xy)^{-1/2}\left(x\frac{dy}{dx} + y\right) = 0$$

But I haven't figured out how to simplify this further. My teacher says that that's all I'll need to know, but I want to understand how the derivative of $\sqrt {xy} = 1$ is $-\frac{y}x$.

Edited to explain that I know the whole thing equals zero, but how do I solve for (dy/dx)?

Attempts to solve get me this far:

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The expression you found should be set equal to $0$ because you differentiated both sides of the equation $\sqrt{xy} = 1$ with respect to $x$. If you have a product of factors equal to $0$, what can you conclude? – Michael Joyce Oct 19 '12 at 13:41
How is $\sqrt{|z|}=1$ different from $|z|=1$? – Thomas Andrews Oct 19 '12 at 13:42
Your subject has $\sqrt{xy}$ while the body of the question has $\sqrt{|xy|}$. Which do you mean? – Thomas Andrews Oct 19 '12 at 13:47

You need to take the derivative of the righthand side of $\sqrt{xy}=1$ as well: the derivative of the constant $1$ is $0$, so you get

$$\frac12(xy)^{-1/2}\left(x\frac{dy}{dx} + y\right)=0\;.$$

Now solve this equation for $\dfrac{dy}{dx}$.

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The question is somewhat unclear, but if $\sqrt{x y}=1$ then $xy=1$, hence $y=x^{-1}$ and $y'=-x^{-2}=-\frac yx=-y^2$.

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You've gotten rid of the absolute value sign how? – Thomas Andrews Oct 19 '12 at 13:46
Differentiability is a local property. If $x>0$ then $y=y(x)>0$, and if $x<0$ then $y=y(x)<0$. The absolute value is not very important, in this question. – Siminore Oct 19 '12 at 14:55
@ThomasAndrews: I got rid of the absolute value just as the OP did mid-post. And apparently Brian made the same simplification. Actually, from $\sqrt{|xy|}=1$ for some function $y\colon x\mapsto y(x)$ we could not even infer that $y$ is a continuous function at all, even less differentiable. – Hagen von Eitzen Oct 19 '12 at 17:35
I like Hagen's solution – Stefan Smith Oct 19 '12 at 23:39

Since no one else has done it, let me expand on Michael Joyce's comment. Since

$$\frac12(xy)^{-\frac12}(x\frac{dy}{dx}+y)=0$$

then either

$$(xy)^{-\frac12}=0 \text{ or } x\frac{dy}{dx}+y=0$$

But if $(xy)^\frac12=1$, then $(xy)^{-\frac12}=1$. Therefore, it follows that

$$x\frac{dy}{dx}+y=0,\frac{dy}{dx}=-\frac yx$$

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