Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\sqrt{|xy|} = 1$

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: enter image description here

share|cite|improve this question
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 $\dfrac{dy}{dx}$.

share|cite|improve this answer

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

share|cite|improve this answer
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


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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.