Implicitly differentiate $sqrt(x+y) = 3x$ without squaring both sides first?

Question

I was just curious about how I could implicitly differentiate $sqrt(x+y) = 3x$ without squaring both sides first. Obviously, if I square both sides first, it becomes "easier" to differentiate and I get:

$dy/dx = 18x-1$

However, whenever I try and implicitly differentiate without squaring both sides first, I get

$dy/dx = 6/(x+y)^{-1/2} -1$

Why is this?

Answer 1

Calling $F(x,y) = \sqrt{x+y} - 3x = 0$ . Then

$$ \frac{dy}{dx} = -\frac{\partial F /\partial x}{\partial F /\partial y}=- \frac{ \frac{1}{2}(x+y)^{-1/2}-3}{\frac{1}{2}(x+y)^{-1/2}}= -1 + 6(x+y)^{1/2} $$

Replacing $(x+y)^{1/2}=3x$ we get

$$ \frac{dy}{dx} = 18x-1$$

Answer 2

You would have $$ \frac{d}{dx}(\sqrt{x + y}) = \frac{1}{2}(x + y)^{-\frac{1}{2}}\frac{d}{dx}(x + y). $$