Question

If $x^3-y^3+3x{y}^2-3x^{2}y+1=0$, then at $(0,1)$ $\frac{dy}{dx}$ equals

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is B $-1$

$x^3-y^3+3x{y}^2-3x^{2}y+1=0$
$\frac{dy}{dx}=\frac{\frac{\delta f}{\delta x}}{\frac{\delta f}{\delta y}}$
Substitute $f=x^3-y^3+3x{y}^2-3x^2+1$
$\left(\frac{\delta f}{\delta x}\right)=3x^2+3y^2-6x$
${\left(\frac{\delta f}{\delta x}\right)}_{(0,1)}=3$
Again $f=x^3-y^3+3x{y}^2-3x^2-3x^{2}y+1$
$\left(\frac{\delta f}{\delta y}\right)=-3y^2+6xy-6x$
${\left(\frac{\delta f}{\delta y}\right)}_{(0,1)}=-3$

So, $\frac{dy}{dx}=\frac{\delta f/\delta x}{\delta f/\delta y}=\frac{3}{-3}=-1$