If $x^2+xy+y^2=\frac{7}{4}$, then $\frac{dy}{dx}$ at x = 1 and $y=\frac{1}{2}$
Answer & explanation
Correct answer: option 2
$x^2+xy+y^2=\frac{7}{4}⇒2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0⇒\frac{dy}{dx}=\frac{-(2x+y)}{x+2y}=\frac{-(2+\frac{1}{2})}{1+1}=\frac{-5}{4}$