# How can I implicitly differentiate (x-y)^3=0?

Apr 20, 2015

If you insist on using implicit differentiation, you'd write:

${\left(x - y\right)}^{3} = 0$

$\frac{d}{\mathrm{dx}} \left({\left(x - y\right)}^{3}\right) = \frac{d}{\mathrm{dx}} \left(0\right)$

$3 {\left(x - y\right)}^{2} \frac{d}{\mathrm{dx}} \left(x - y\right) = 0$

$3 {\left(x - y\right)}^{2} \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

So $3 {\left(x - y\right)}^{2} - 3 {\left(x - y\right)}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

And $3 {\left(x - y\right)}^{2} = 3 {\left(x - y\right)}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

And ${\left(x - y\right)}^{2} = {\left(x - y\right)}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$.

Now we can say, either $x - y = 0$, or we can divide by $x - y$ to get $\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

The difficulty is that the only solutions to ${\left(x - y\right)}^{3} = 0$ are $x - y = 0$
so $y = x$ and $\frac{\mathrm{dy}}{\mathrm{dx}} = 1$. Why use implicit differentiation?

Apr 20, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Rewrite it as ${x}^{3} - 3 {x}^{2} y + 3 x {y}^{2} - {y}^{3}$=0

Now differentiate term by term with respect to x,

$3 {x}^{2} - 6 x y - 3 {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} + 6 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$=0

$- 3 \left({x}^{2} - 2 x y + {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} + 6 x y - 3 {y}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$