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:
3 Answers
$\begingroup$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 $\dfrac{dy}{dx}$.
$\endgroup$ $\begingroup$The question is somewhat unclear, but if $\sqrt{x y}=1$ then $xy=1$, hence $y=x^{-1}$ and $y'=-x^{-2}=-\frac yx=-y^2$.
$\endgroup$ 4 $\begingroup$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
$$x\frac{dy}{dx}+y=0,\frac{dy}{dx}=-\frac yx$$
$\endgroup$
