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

Continuity and Differentiability

Solution:

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