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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?

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  • $\begingroup$ I'm not sure how you are getting $18x - 1$. The answer should be $-1$. $\endgroup$ – wgrenard Oct 8 '16 at 21:05
  • $\begingroup$ @wgrenard poop, I mistyped the equation. It's supposed to be 3x on the right hand side. $\endgroup$ – ZERO Oct 8 '16 at 21:06
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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

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

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

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  • $\begingroup$ Yeah I see that and then when you simplify $d/dx(x+6)$ you get $dy/dx$. Then you get $dy/dx=6/(x+y)^.5$ when you simplify the whole equation, as I mentioned in the OP. $\endgroup$ – ZERO Oct 8 '16 at 21:05
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    $\begingroup$ @ZERO You should get $y' = 6\sqrt{x+y} - 1$. Which is the same thing as $18x - 1$ since $\sqrt{x+y} = 3x$ $\endgroup$ – wgrenard Oct 8 '16 at 21:10
  • $\begingroup$ Oh wow, that makes sense, thanks! $\endgroup$ – ZERO Oct 8 '16 at 21:19

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