Application of derivative - to find interval in which abscissa changes faster than then ordinate Problem : On the curve $x^3=12y$ find the interval of values of x for which the abscissa changes at a faster rate than the ordinate ? Solution : We differentiate w.r.t. to y we get : $3x^2\frac{dx}{dy} = 12$ $\frac{dx}{dy} = \frac{12}{3x^2}$ Now to find the interval in which the abscissa changes at a faster rate than the ordinate, we must have : $|\frac{dx}{dy}| >1$ I am unable to understand this point.. please explain...this point thanks...

Note: I am simply following the work the OP has already done

if you look at

$$ \frac{\mathrm{d}x}{\mathrm{d}y}=\frac{12}{3x^2} $$

then you'll be able to see that

$$ \left\lvert \frac{\mathrm{d}x}{\mathrm{d}y} \right\rvert = \frac{12}{3x^2} = 1 \text{ when } x = \pm 2 $$

Looking outside the interval $[-2,2]$ we can see $\left\lvert \frac{\mathrm{d}x}{\mathrm{d}y} \right\rvert < 1$ (the denominator is clearly larger than the numerator) so all that's left is inside the interval. We can see that if you plug in anything side the interval the inequality holds (except for at $x=0$ which it is not defined). So it appears to be to be $[-2,0) \cup (0,2]$