Implicit Differentiation of $2\sqrt{x} + \sqrt{y} = 3$

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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 $$\dfrac{dy}{dx} = 4 - 12 \dfrac12 \dfrac1{\sqrt{x}} = 4 - \dfrac{6}{\sqrt{x}}$$