Proving/disproving that $1=\frac {1}{2}(x)^{-\frac {1}{2}} \Longleftrightarrow 2\sqrt{x}=0$ I was looking at the solution to this problem:
Find $\frac {dy}{dx}$ for $x=10+\sqrt {x^2+y^2}$.
The solution given was as follows:
$$
\frac {d}{dx}(x)=\frac {d}{dx}(10)+\frac {d}{dx}(\sqrt {x^2+y^2})
$$
$$
1=\frac {1}{2}(x^2+y^2)^{-\frac {1}{2}}(2x+2y\frac {dy}{dx})
$$
$$
2\sqrt{x^2+y^2}=2x+2y\frac {dy}{dx}
$$
$$
2\sqrt{x^2+y^2}-2x=+2y\frac {dy}{dx}
$$
$$
\frac {dy}{dx}=\frac {\sqrt{x^2+y^2}-x}{y}
$$
I understand it except for what happens between step two and three. I tried but was unable to prove to myself that
$$
1=\frac {1}{2}(x)^{-\frac {1}{2}} \Longleftrightarrow 2\sqrt{x}=0
$$
or that (I think this is equivalent)
$$
-(\frac {1}{2}(x)^{-\frac {1}{2}})+1=2\sqrt{x}
$$
Can someone show me how to prove it, or am I looking at the original problem wrong? Thank you.

Rewriting from step 2 might help:$$1=\frac {1}{2}(x^2+y^2)^{-\frac {1}{2}}(2x+2y\frac {dy}{dx})$$$$1=\frac {1}{2\sqrt{x^2+y^2}}(2x+2y\frac {dy}{dx})$$$$1=\frac{(2x+2y\frac {dy}{dx})}{2\sqrt{x^2+y^2}}$$Now just multiply both sides by $2\sqrt{x^2+y^2}$
