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

1 Answer
Aug 10, 2017

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