Question

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

Answer 1

We have:

`y^2 = (x-a)^2(x-b)`

Using the product rule, and differentiating implicitly:

`2y(dy/dx) = (x-a)^2(1) + 2(x-a)(x-b)`
`" " = (x-a)(x-a +2x-2b)`
`" " = (x-a)(3x-a -2b)`

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

`(2y)( (d^2y)/(dx^2) ) + (2dy/dx)(dy/dx) = (x-a)(3) + (1)(3x-a -2b)`

So when the second derivative vanishes we get the equation

`2(dy/dx)^2 = 3(x-a) + (3x-a -2b)`
`" " = 3x-3a + 3x-a -2b`
`" " = 6x-4a -2b`

Substituting for `dy/dx`

`2(((x-a)(3x-a -2b))/(2y))^2 = 6x-4a -2b`

`:. (2 (x-a)^2(3x-a -2b)^2 )/(4y^2) = 6x-4a -2b`

`:. ( (x-a)^2(3x-a -2b)^2 )/(2y^2) = 6x-4a -2b`

`:. ( (x-a)^2(3x-a -2b)^2 )/((x-a)^2(x-b)) = 2(6x-4a -2b)`

`:. ( (3x-a -2b)^2 )/((x-b)) = 2(6x-4a -2b)`

`:. (3x-a -2b)^2 = 2(6x-4a -2b)(x-b)`

`:. a^2+4ab-6ax+4b^2-12bx+9x^2 = 8ab-8ax+4b^2-16bx+12x^2`

`:. 8ab-8ax+4b^2-16bx+12x^2 -a^2-4ab+6ax-4b^2+12bx-9x^2 = 0`

`:. 4ab-2ax-4bx+3x^2 -a^2 = 0`

`:. 3x^2 - (2a+4b)x + 4ab -a^2 = 0`

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

`x = ( (2a+4b) +- sqrt( (- (2a+4b))^2 - 4(3)(4ab-a^2) ) ) / 6`
`\ \ = ( (2a+4b) +- sqrt( 4a^2+16ab+16b^2 -48ab+ 12a^2 ) ) / 6`
`\ \ = ( (2a+4b) +- sqrt( 16a^2-32ab+16b^2 ) ) / 6`
`\ \ = ( (2a+4b) +- sqrt( 16(a^2-2ab+b^2) ) ) / 6`
`\ \ = ( (2a+4b) +- sqrt( 16(a-b)^2 ) ) / 6`
`\ \ = ( (2a+4b) +- 4 (a-b) ) / 6`

So the two roots are:

`x_1 = ( (2a+4b) - 4 (a-b) ) / 6`
`\ \ \ = ( 2a+4b -4a+4b ) / 6`
`\ \ \ = ( -2a+8b ) / 6`
`\ \ \ = ( 4b - a ) / 3`

Or:

`x_2 = ( (2a+4b) + 4 (a-b) ) / 6`
`\ \ = ( 2a+4b +4a-4b ) / 6`
`\ \ = ( 6a ) / 6`
`\ \ = a`

So one point of inflection is:

`x=a`

And others satisfy:

`x= ( 4b - a ) / 3 => 3x = 4b-a`
`:. 3x+4b=a \ \ \` QED