Find \; d y/d x for x² y²+3xy+y=0

Correct Answer is : − 2xy2+3y2x2y+3x+1-\; {2xy²+3y}{2x²y+3x+1}−2x2y+3x+12xy2+3y

Correct Answer is : − 2xy2+3y2x2y+3x+1-\; {2xy²+3y}{2x²y+3x+1}−2x2y+3x+12xy2+3y

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Differential Calculus

Question : 52 of 69

Marks: +1, -0

Find

\; \frac{d y}{d x}

x^{2} y^{2}+3xy+y=0

Solution:

Concept:

Use implicit differentiation: differentiate both sides with respect to

x

, treating

y

as a function of

x

Explanation:

Differentiate each term of

x^{2}y^{2}+3xy+y=0

•

\frac{d}{dx}(x^{2}y^{2}) = 2x y^{2} + x^{2} \cdot 2y \frac{dy}{dx} = 2xy^{2} + 2x^{2}y\frac{dy}{dx}

•

\frac{d}{dx}(3xy) = 3y + 3x\frac{dy}{dx}

•

\frac{d}{dx}(y) = \frac{dy}{dx}

Sum:

2xy^{2} + 2x^{2}y\frac{dy}{dx} + 3y + 3x\frac{dy}{dx} + \frac{dy}{dx} = 0

Collect

\frac{dy}{dx}

terms:

(2x^{2}y + 3x + 1)\frac{dy}{dx} + (2xy^{2}+3y) = 0

Solve:

\frac{dy}{dx} = -\frac{2xy^{2}+3y}{2x^{2}y+3x+1}

Answer:

\frac{dy}{dx} = -\frac{2xy^{2}+3y}{2x^{2}y+3x+1}

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