Miscellaneous · Q14: Solve $x\sqrt{1+y}+y\sqrt{1+x}=0$ as $y=-\frac{x}{1+x}$⁠, so $\frac{dy}{dx}=-\frac{1}{(1+x)^2}$

Basics you need

Solution

1. Isolate one term and square

   Move the second term to the right-hand side:

   $$x\sqrt{1+y} = -y\sqrt{1+x}.$$

   Squaring both sides (and using $(\sqrt{1+y})^2 = 1+y$⁠, $(\sqrt{1+x})^2 = 1+x$⁠):

   $$x^2(1+y) = y^2(1+x).$$
   Each side now has a single square root, so squaring removes both roots at once. Squaring is allowed (equal numbers have equal squares), but it may let in extra solutions where $x\sqrt{1+y} = +y\sqrt{1+x}$ instead. We deal with that in Step 3.
   $(ab)^2 = a^2b^2$ and $(\sqrt u)^2 = u$ for $u \ge 0$ (Basic 1).

2. Factorise the squared equation

   $$\begin{aligned} & x^2 + x^2y - y^2 - xy^2 = 0 \\ \Longrightarrow\ & (x^2 - y^2) + xy(x - y) = 0 \\ \Longrightarrow\ & (x - y)(x + y) + xy(x - y) = 0 \\ \Longrightarrow\ & (x - y)(x + y + xy) = 0. \end{aligned}$$

   So $y = x$⁠, or $x + y + xy = 0$⁠, that is $y(1 + x) = -x$⁠, i.e.

   $$y = -\frac{x}{1+x} \qquad (1 + x \ne 0 \text{ since } x > -1).$$
   Grouping $x^2 - y^2$ and $x^2y - xy^2 = xy(x-y)$ exposes the common factor $x - y$⁠. A product is zero exactly when one of its factors is zero, so there are only two candidates for $y$⁠.
   Difference of squares and common factors (Basic 2). Basics: solving for dy/dx

3. Test both candidates in the original equation

   Candidate $y = x$⁠. The left side becomes $x\sqrt{1+x} + x\sqrt{1+x} = 2x\sqrt{1+x}$⁠. For $-1 < x < 1$⁠, $\sqrt{1+x} > 0$⁠, so this is 0 only when $x = 0$⁠. So $y = x$ is not a solution except at the single point $(0,0)$⁠.

   Candidate $y = -\frac{x}{1+x}$⁠. First, $1 + y = \frac{1 + x - x}{1+x} = \frac{1}{1+x} > 0$⁠, so $\sqrt{1+y} = \frac{1}{\sqrt{1+x}}$ exists. Then

   $$\begin{aligned} x\sqrt{1+y} + y\sqrt{1+x} &= \frac{x}{\sqrt{1+x}} - \frac{x\sqrt{1+x}}{1+x} \\ &= \frac{x}{\sqrt{1+x}} - \frac{x}{\sqrt{1+x}} = 0. \ ✓ \end{aligned}$$

   Hence the points satisfying the given equation with $-1 < x < 1$ are exactly the points of the curve

   $$y = -\frac{x}{1+x}, \qquad -1 < x < 1.$$
   This is the step that squaring forces on us. The line $y = x$ satisfies the squared equation but not the original one (try $x = 0.5$⁠: the left side is $2(0.5)\sqrt{1.5} \ne 0$⁠). Its only point that does satisfy the original, $(0,0)$⁠, already lies on the curve $y = -\frac{x}{1+x}$⁠. So the given equation defines $y$ as this one function of $x$⁠, and that is the function we differentiate. (A common shortcut, dividing the squared equation by $x - y$⁠, needs exactly this justification.)
   Using $\frac{\sqrt{1+x}}{1+x} = \frac{1}{\sqrt{1+x}}$ for $1 + x > 0$⁠. Checking solutions after squaring (Basic 1).

4. Differentiate with the quotient rule

   With $u = x$⁠, $v = 1 + x$⁠, $u' = 1$⁠, $v' = 1$⁠, and $v \ne 0$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= -\frac{u'v - uv'}{v^2} \\ &= -\frac{1\cdot(1+x) - x\cdot 1}{(1+x)^2} \\ &= -\frac{1}{(1+x)^2}. \end{aligned}$$

   This holds for every $x$ with $-1 < x < 1$⁠, as required.

   The minus sign in front stays outside the fraction. The numerator simplifies to $1 + x - x = 1$⁠, which is why the answer is so clean.
   Quotient rule, rule (3) of Section 5.3 (Basic 3). Basics: derivative rules

5. Confirm by implicit differentiation

   Differentiate the original equation term by term with respect to $x$⁠, using the product rule and the chain rule $\frac{d}{dx}\sqrt{1+y} = \frac{1}{2\sqrt{1+y}}\frac{dy}{dx}$⁠:

   $$\begin{aligned} & \sqrt{1+y} + \frac{x}{2\sqrt{1+y}}\frac{dy}{dx} \\ & {}+ \frac{dy}{dx}\sqrt{1+x} + \frac{y}{2\sqrt{1+x}} = 0. \end{aligned}$$

   Collect the $\frac{dy}{dx}$ terms:

   $$\frac{dy}{dx} = -\,\frac{\sqrt{1+y} + \dfrac{y}{2\sqrt{1+x}}}{\sqrt{1+x} + \dfrac{x}{2\sqrt{1+y}}}.$$

   Now substitute $\sqrt{1+y} = \frac{1}{\sqrt{1+x}}$ and $y = -\frac{x}{1+x}$⁠, and write $s = \sqrt{1+x}$ (so $s^2 = 1 + x$⁠). The numerator is $\frac{1}{s} - \frac{x}{2s^3} = \frac{2s^2 - x}{2s^3} = \frac{2 + x}{2s^3}$⁠, and the denominator is $s + \frac{xs}{2} = \frac{s(2 + x)}{2}$⁠. Dividing,

   $$\frac{dy}{dx} = -\frac{2+x}{2s^3}\cdot\frac{2}{s(2+x)} = -\frac{1}{s^4} = -\frac{1}{(1+x)^2}. \ ✓$$
   The differentiation itself never uses the solved form of $y$ (only the final simplification does), so it checks the quotient-rule work of Step 4 by a different route. Cancelling $2 + x$ is fine because $x > -1$ makes $2 + x > 1$⁠. The denominator of the implicit formula is not zero, so the formula is valid at every point of the curve.
   Implicit differentiation, Section 5.3.2; product rule (rule (2), Section 5.3); chain rule, Theorem 4. Lesson: implicit differentiation

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