If $x \sqrt{1+y} + y \sqrt{1+x} = 0$, for $-1 < x < 1$, prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$

If x √(1+y) + y √(1+x) = 0, for -1 < x < 1, prove – Class 12 Mathematics

Miscellaneous Exercise · Q14

x \sqrt{1+y} + y \sqrt{1+x} = 0

Tripura TbseTextbookSubjective· 3mImportance★★★★★

Appeared in past exams:MHT-CET 2020· Set pcm-2020-10-16-M· 2mreworded

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The relation reduces to $y=-\frac{x}{1+x}$ (the valid branch), and differentiating it gives $\frac{dy}{dx}=-\frac{1}{(1+x)^2}$ , as required.

Rather than differentiate the messy square-root equation directly, we first uncover the simple relationship it hides.

Step 1 — rewrite and square

x\sqrt{1+y} + y\sqrt{1+x} = 0 \;\Rightarrow\; x\sqrt{1+y} = -y\sqrt{1+x}.

Square both sides (we will check for extraneous roots afterward):

x^2(1+y) = y^2(1+x).

Step 2 — factor

x^2 - y^2 + x^2y - xy^2 = 0 \;\Rightarrow\; (x-y)(x+y) + xy(x-y) = 0 \;\Rightarrow\; (x-y)(x+y+xy) = 0.

So either $x-y=0$ or $x+y+xy=0$ .

Step 3 — pick the correct branch

If $y=x$ , the original equation becomes $2x\sqrt{1+x}=0$ , true only at $x=0$ — not a valid relation on the whole interval, so it is extraneous from squaring. Therefore we take

$x+y+xy = 0 \;\Rightarrow\; y = -\frac{x}{1+x}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2026Set may-20261 markMCQ
$x^y = y^x$
›Reveal solutionSolution

Log-differentiate $x^y=y^x$ : from $y\log x=x\log y$ you get $\dfrac{dy}{dx}=\dfrac{y(x\log y-y)}{x(y\log x-x)}$ .

Step 1 — Take logarithms of both sides
$x^y=y^x\ \Rightarrow\ y\log x=x\log y$
Step 2 — Differentiate implicitly with respect to $x$

Apply the product rule to each side:
$y'\log x+\frac{y}{x}=\log y+x\cdot\frac{y'}{y}$
Step 3 — Collect the $y'$ terms
$y'\log x-\frac{x\,y'}{y}=\log y-\frac{y}{x}$ $y'\left(\log x-\frac{x}{y}\right)=\log y-\frac{y}{x}$
Step 4 — Solve for $y'$ and tidy the fractions

$y'=\frac{\log y-\dfrac{y}{x}}{\log x-\dfrac{x}{y}}=\frac{\dfrac{x\log y-y}{x}}{\dfrac{y\log x-x}{y}}=\frac{y\,(x\log y-y)}{x\,(y\log x-x)}$ …
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CA Foundation 2025Set sep-20251 markMCQ
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Implicit differentiation of $x^2y^2+y=0$ gives $\dfrac{dy}{dx} = \dfrac{-2xy^2}{2x^2y+1}$ .

Step 1 — Differentiate term by term (y depends on x)

For $x^2y^2$ use the product rule:
$\frac{d}{dx}(x^2y^2) = 2x\,y^2 + x^2\cdot 2y\frac{dy}{dx} = 2xy^2 + 2x^2y\frac{dy}{dx}$
and $\dfrac{d}{dx}(y) = \dfrac{dy}{dx}$ .

Step 2 — Assemble the differentiated equation
$2xy^2 + 2x^2y\frac{dy}{dx} + \frac{dy}{dx} = 0$
Step 3 — Collect and solve for dy/dx
$\frac{dy}{dx}\big(2x^2y + 1\big) = -2xy^2$ $\frac{dy}{dx} = \frac{-2xy^2}{2x^2y + 1}$
Why the other options are wrong: (A) drops the minus sign; (C) and (D) wrongly move the "+1" into the numerator, which happens only if you fail to factor $\frac{dy}{dx}$ correctly. …
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←Previous question · misc-exerciseQ13 — Find (dy)/(dx), if y = sin⁻¹ x + sin⁻¹ √(1-x²), 0 < x < 1 →Next question · misc-exerciseQ15 — If (x-a)² + (y-b)² = c², for some c > 0, prove that frac[1+((dy)/(dx))…

Q.If $x \sqrt{1+y} + y \sqrt{1+x} = 0$ , for $-1 < x < 1$ , prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$

Concept understanding — Implicit Differentiation

Implicit Differentiation

When $y$ isn't alone

The one key move: $y$ is really $y(x)$

The procedure

Worked example

Step 1 — rewrite and square

Step 2 — factor

Step 3 — pick the correct branch

Method: Simplify the Implicit Relation Before Differentiating

Steps

Common Mistakes

Step 1 — Take logarithms of both sides

Step 2 — Differentiate implicitly with respect to $x$

Step 3 — Collect the $y'$ terms

Step 4 — Solve for $y'$ and tidy the fractions

Step 1 — Differentiate term by term (y depends on x)

Step 2 — Assemble the differentiated equation

Step 3 — Collect and solve for dy/dx

Q.If x1+y+y1+x=0x \sqrt{1+y} + y \sqrt{1+x} = 0x1+y​+y1+x​=0, for −1<x<1-1 < x < 1−1<x<1, prove that dydx=−1(1+x)2\frac{dy}{dx} = -\frac{1}{(1+x)^2}dxdy​=−(1+x)21​

Step 1 — rewrite and square

Step 2 — factor

Step 3 — pick the correct branch

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Q.If $x \sqrt{1+y} + y \sqrt{1+x} = 0$ , for $-1 < x < 1$ , prove that $\frac{dy}{dx} = -\frac{1}{(1+x)^2}$

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