If $x^2+xy+y^2=\frac{7}{4}$, then $\frac

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Continuity and Differentiability

Question:

If $x^2+xy+y^2=\frac{7}{4}$, then $\frac{dy}{dx}$ at x = 1 and $y=\frac{1}{2}$

Options:

$\frac{3}{4}$

$\frac{-5}{4}$

$\frac{21}{8}$

$\frac{-21}{8}$

Correct Answer:

$\frac{-5}{4}$

Explanation:

$x^2+xy+y^2=\frac{7}{4}⇒2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0⇒\frac{dy}{dx}=\frac{-(2x+y)}{x+2y}=\frac{-(2+\frac{1}{2})}{1+1}=\frac{-5}{4}$