If $(x^2 + y^2)^2 = xy$, find $\dfrac{dy}{dx}$. **OR** If $x = a(2\theta - \sin 2\theta)$ and $y = a(1 - \cos 2\theta)$, find $\dfrac{dy}{dx}$ when $\theta = \dfrac{\pi}{3}$.

If (x² + y²)² = xy, find (dy)/(dx). **OR** If x = – Class 12 Mathematics

Question 280 of 281

(x^2 + y^2)^2 = xy

Uttar Pradesh UpmspCBSE Class XII Board 2018Subjective· 4mImportance★★★★★

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$\dfrac{dy}{dx}=\dfrac{y-4x(x^2+y^2)}{4y(x^2+y^2)-x}$ (and the OR case gives $\dfrac{1}{\sqrt3}$ at $\theta=\tfrac\pi3$ ).

Concept. Implicit differentiation with the chain and product rules.

Why this method. $y$ is defined implicitly, so we differentiate both sides in $x$ and collect $\dfrac{dy}{dx}$ .

Working. From $(x^2+y^2)^2=xy$ :

$2(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)=y+x\frac{dy}{dx}.$

$4(x^2+y^2)x+4(x^2+y^2)y\frac{dy}{dx}=y+x\frac{dy}{dx}.$

$\frac{dy}{dx}\big[4y(x^2+y^2)-x\big]=y-4x(x^2+y^2).$

$\boxed{\ \frac{dy}{dx}=\frac{y-4x(x^2+y^2)}{4y(x^2+y^2)-x}\ }$

OR (parametric). $x=a(2\theta-\sin2\theta),\ y=a(1-\cos2\theta)$ : …

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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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Q.If $(x^2 + y^2)^2 = xy$ , find $\dfrac{dy}{dx}$ . OR If $x = a(2\theta - \sin 2\theta)$ and $y = a(1 - \cos 2\theta)$ , find $\dfrac{dy}{dx}$ when $\theta = \dfrac{\pi}{3}$ .

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 — 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 (x2+y2)2=xy(x^2 + y^2)^2 = xy(x2+y2)2=xy, find dydx\dfrac{dy}{dx}dxdy​. OR If x=a(2θ−sin⁡2θ)x = a(2\theta - \sin 2\theta)x=a(2θ−sin2θ) and y=a(1−cos⁡2θ)y = a(1 - \cos 2\theta)y=a(1−cos2θ), find dydx\dfrac{dy}{dx}dxdy​ when θ=π3\theta = \dfrac{\pi}{3}θ=3π​.

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Q.If $(x^2 + y^2)^2 = xy$ , find $\dfrac{dy}{dx}$ . OR If $x = a(2\theta - \sin 2\theta)$ and $y = a(1 - \cos 2\theta)$ , find $\dfrac{dy}{dx}$ when $\theta = \dfrac{\pi}{3}$ .

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