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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$\begingroup$ Could you do it if it were $2x^2+y^2=3$? The mechanics are exactly the same. $\endgroup$ – Brian M. Scott Jun 29 '12 at 21:57
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1$\begingroup$ What is the problem with the square root? Do you know the answer to $\partial_x \sqrt{x}$? $\endgroup$ – Fabian Jun 29 '12 at 21:58
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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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$\begingroup$ 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? $\endgroup$ – Yai0Phah Jun 29 '12 at 22:59
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$\begingroup$ Good question, and I do not know the answer to that, sorry. $\endgroup$ – Joe Jun 29 '12 at 23:02
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Implicit diff works. You have $${1\over\sqrt{x}} + {y'\over 2\sqrt{y}} = 0.$$ Now solve for $y'$.
answered Jun 29 '12 at 21:58
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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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$\begingroup$ I thought about doing this, but the poser said "implicit diff" so I complied with the request. +1 for a good solution. $\endgroup$ – ncmathsadist Jun 30 '12 at 0:10
