Question

If $y=\sqrt[3]{6x{y}^2-11x^{2}y+6x^3}$, then possible value(s) of $\frac{dy}{dx}$ at $x=2$ is(are)

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B), (C), (D)

$y=\sqrt[3]{6x{y}^2-11x^{2}y+6x^3}$
$y^3=6x{y}^2-11x^{2}y+6x^3$
$y^3-6x{y}^2+11x^{2}y-6x^3=0$
$\Rightarrow (y-x)(y-2x)(y-3x)=0$
$\Rightarrow y=x,y=2x,y=3x$
$\therefore \frac{dy}{dx}=1,2,3$