Basic Applications of CalculusPYQ Sept 25Question 4136 of 28
All Questions

Find dydx\displaystyle \frac{dy}{dx} for x2y2+y=0\displaystyle x^2y^2+y = 0.

Options

A2y2x2y2x2+1\displaystyle \frac{2y^2x}{2y^2x^2+1}
B−2y2x2yx2+1\displaystyle \frac{-2y^2x}{2yx^2+1}
C−2y2x+12y2x2\displaystyle \frac{-2y^2x+1}{2y^2x^2}
D2y2x−12y2x2\displaystyle \frac{2y^2x-1}{2y^2x^2}
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Correct Answer

✅ Option b — −2y2x2yx2+1\displaystyle \frac{-2y^2x}{2yx^2+1}

All Options:

  • A2y2x2y2x2+1\displaystyle \frac{2y^2x}{2y^2x^2+1}
  • B−2y2x2yx2+1\displaystyle \frac{-2y^2x}{2yx^2+1}
  • C−2y2x+12y2x2\displaystyle \frac{-2y^2x+1}{2y^2x^2}
  • D2y2x−12y2x2\displaystyle \frac{2y^2x-1}{2y^2x^2}

Detailed Solution & Explanation

Given equation: x2y2+y=0x^2y^2 + y = 0
We differentiate both sides with respect to x\displaystyle x using implicit differentiation: ddx(x2y2+y)=ddx(0)\frac{d}{dx}\left(x^2y^2 + y\right) = \frac{d}{dx}(0)
Using the product rule on the first term: [ddx(x2)⋅y2+x2⋅ddx(y2)]+dydx=0\left[ \frac{d}{dx}(x^2) \cdot y^2 + x^2 \cdot \frac{d}{dx}(y^2) \right] + \frac{dy}{dx} = 0 [2xy2+x2(2ydydx)]+dydx=0\left[ 2xy^2 + x^2 \left(2y \frac{dy}{dx}\right) \right] + \frac{dy}{dx} = 0 2xy2+2x2ydydx+dydx=02xy^2 + 2x^2y\frac{dy}{dx} + \frac{dy}{dx} = 0
Collect the terms with dydx\displaystyle \frac{dy}{dx}: dydx(2x2y+1)=−2xy2\frac{dy}{dx}\left(2x^2y + 1\right) = -2xy^2 Solve for dydx\displaystyle \frac{dy}{dx}: dydx=−2xy22x2y+1=−2y2x2yx2+1\frac{dy}{dx} = \frac{-2xy^2}{2x^2y + 1} = \frac{-2y^2x}{2yx^2+1}
Hence, **Option B** is the correct answer.

About This Chapter: Basic Applications of Calculus

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Limits, Continuity, Derivatives, Integrals

This chapter covers Limits, Continuity, Derivatives, Integrals and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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