# How do you differentiate  y^2 = (x-a)^2(x-b)  implicitly?

Aug 10, 2017

We have:

${y}^{2} = {\left(x - a\right)}^{2} \left(x - b\right)$

Using the product rule, and differentiating implicitly:

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

For a point of inflection we look to see where the second derivative vanishes, so let us differentiate again:

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

So when the second derivative vanishes we get the equation

$2 {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 3 \left(x - a\right) + \left(3 x - a - 2 b\right)$
$\text{ } = 3 x - 3 a + 3 x - a - 2 b$
$\text{ } = 6 x - 4 a - 2 b$

Substituting for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$2 {\left(\frac{\left(x - a\right) \left(3 x - a - 2 b\right)}{2 y}\right)}^{2} = 6 x - 4 a - 2 b$

$\therefore \frac{2 {\left(x - a\right)}^{2} {\left(3 x - a - 2 b\right)}^{2}}{4 {y}^{2}} = 6 x - 4 a - 2 b$

$\therefore \frac{{\left(x - a\right)}^{2} {\left(3 x - a - 2 b\right)}^{2}}{2 {y}^{2}} = 6 x - 4 a - 2 b$

$\therefore \frac{{\left(x - a\right)}^{2} {\left(3 x - a - 2 b\right)}^{2}}{{\left(x - a\right)}^{2} \left(x - b\right)} = 2 \left(6 x - 4 a - 2 b\right)$

$\therefore \frac{{\left(3 x - a - 2 b\right)}^{2}}{\left(x - b\right)} = 2 \left(6 x - 4 a - 2 b\right)$

$\therefore {\left(3 x - a - 2 b\right)}^{2} = 2 \left(6 x - 4 a - 2 b\right) \left(x - b\right)$

$\therefore {a}^{2} + 4 a b - 6 a x + 4 {b}^{2} - 12 b x + 9 {x}^{2} = 8 a b - 8 a x + 4 {b}^{2} - 16 b x + 12 {x}^{2}$

$\therefore 8 a b - 8 a x + 4 {b}^{2} - 16 b x + 12 {x}^{2} - {a}^{2} - 4 a b + 6 a x - 4 {b}^{2} + 12 b x - 9 {x}^{2} = 0$

$\therefore 4 a b - 2 a x - 4 b x + 3 {x}^{2} - {a}^{2} = 0$

$\therefore 3 {x}^{2} - \left(2 a + 4 b\right) x + 4 a b - {a}^{2} = 0$

Which is a quadratic that we can solve using the quadratic formula:

$x = \frac{\left(2 a + 4 b\right) \pm \sqrt{{\left(- \left(2 a + 4 b\right)\right)}^{2} - 4 \left(3\right) \left(4 a b - {a}^{2}\right)}}{6}$
$\setminus \setminus = \frac{\left(2 a + 4 b\right) \pm \sqrt{4 {a}^{2} + 16 a b + 16 {b}^{2} - 48 a b + 12 {a}^{2}}}{6}$
$\setminus \setminus = \frac{\left(2 a + 4 b\right) \pm \sqrt{16 {a}^{2} - 32 a b + 16 {b}^{2}}}{6}$
$\setminus \setminus = \frac{\left(2 a + 4 b\right) \pm \sqrt{16 \left({a}^{2} - 2 a b + {b}^{2}\right)}}{6}$
$\setminus \setminus = \frac{\left(2 a + 4 b\right) \pm \sqrt{16 {\left(a - b\right)}^{2}}}{6}$
$\setminus \setminus = \frac{\left(2 a + 4 b\right) \pm 4 \left(a - b\right)}{6}$

So the two roots are:

${x}_{1} = \frac{\left(2 a + 4 b\right) - 4 \left(a - b\right)}{6}$
$\setminus \setminus \setminus = \frac{2 a + 4 b - 4 a + 4 b}{6}$
$\setminus \setminus \setminus = \frac{- 2 a + 8 b}{6}$
$\setminus \setminus \setminus = \frac{4 b - a}{3}$

Or:

${x}_{2} = \frac{\left(2 a + 4 b\right) + 4 \left(a - b\right)}{6}$
$\setminus \setminus = \frac{2 a + 4 b + 4 a - 4 b}{6}$
$\setminus \setminus = \frac{6 a}{6}$
$\setminus \setminus = a$

So one point of inflection is:

$x = a$

And others satisfy:

$x = \frac{4 b - a}{3} \implies 3 x = 4 b - a$
$\therefore 3 x + 4 b = a \setminus \setminus \setminus$ QED