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I am trying to implicitly differentiate this problem below but I am stumped because of the square-roots. $$2\sqrt{x} + \sqrt{y} = 3$$

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Could you do it if it were $2x^2+y^2=3$? The mechanics are exactly the same. – Brian M. Scott Jun 29 '12 at 21:57
What is the problem with the square root? Do you know the answer to $\partial_x \sqrt{x}$? – Fabian Jun 29 '12 at 21:58
up vote 1 down vote accepted

Our original is:

$$2\sqrt{x} + \sqrt{y} = 3 \tag{1}$$

Taking the derivative with respect to $x$ and recalling that the derivative of a constant is zero, we get:

$$\frac{1}{\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot y' = 0$$

Cross multiply and you get

$$2\sqrt{y} + y'\sqrt{x} = 0$$

Now subtract the $2\sqrt{y}$ term so we are dealing with the $y'$ term on one side only to get:

$$y'\sqrt{x} = -2\sqrt{y}$$

Divide through by $\sqrt{x}$ to get $y'$ by itself and you find that

$$y' = \frac{-2\sqrt{y}}{\sqrt{x}} \tag{2}$$

To show how this is equivalent to the answer provided by Marvis, look at equation $(1)$ and solve for $\sqrt{y}$ directly and you find that:

$$\sqrt{y} = 3 - 2\sqrt{x} \tag{3}$$

Moving along to equation $(2)$ and replacing the $\sqrt{y}$ term with $3-2\sqrt{x}$, we get:

$$y' = \frac{-2\cdot (3-2\sqrt{x})}{\sqrt{x}}$$

$$y' = \frac{-6 + 4\sqrt{x}}{\sqrt{x}}$$

$$y' = \frac{-6}{\sqrt{x}} + 4 \equiv 4 -\frac{6}{\sqrt{x}} \tag{4}$$

Remark: I would argue implicit differentiation is simple enough here and I would stop at $(2)$. If needed to simplify further though, I would have gone the route Marvis took by rewriting the original and squaring to find $y = \text{something}$ and then taking the derivative of both sides.

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Why is it usually not checked whether $\partial F/\partial x$ and $\partial F/\partial y$ are continuous in some neighborhood of $(x_0,y_0)$, and $\partial F/\partial y\neq0$ at $(x_0,y_0)$ when we are calculating the implicit differentiation? – Frank Science Jun 29 '12 at 22:59
Good question, and I do not know the answer to that, sorry. – Joe Jun 29 '12 at 23:02

Implicit diff works. You have $${1\over\sqrt{x}} + {y'\over 2\sqrt{y}} = 0.$$ Now solve for $y'$.

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You could also get away from implicit differentiation for this problem as follows. Since $2 \sqrt{x} + \sqrt{y} = 3$, we get that $\sqrt{y} = 3 - 2 \sqrt{x} \implies y = (3-2\sqrt{x})^2 = 9 + 4x - 12 \sqrt{x}$. Now you can differentiate it to get $$\dfrac{dy}{dx} = 4 - 12 \dfrac12 \dfrac1{\sqrt{x}} = 4 - \dfrac{6}{\sqrt{x}}$$

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I thought about doing this, but the poser said "implicit diff" so I complied with the request. +1 for a good solution. – ncmathsadist Jun 30 '12 at 0:10

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