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

Question

Cameron Alexander · Accepted Answer

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

Charles Anctil · Answer

undefined To implicitly differentiate $y^2 = (x-a)^2(x-b)$:

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the chain rule where necessary.
3. Use the product rule when differentiating terms that involve products of functions.
4. Solve for $\frac{dy}{dx}$.

Differentiating each term:

$$\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$$

$$\frac{d}{dx}((x-a)^2(x-b)) = 2(x-a)(x-b) + (x-a)^2\frac{d}{dx}(x-b) + (x-b)\frac{d}{dx}((x-a)^2)$$

$$= 2(x-a)(x-b) + (x-a)^2 - (x-b)(2(x-a))$$

Putting it all together and solving for $\frac{dy}{dx}$:

$$2y\frac{dy}{dx} = 2(x-a)(x-b) + (x-a)^2 - 2(x-a)(x-b) - (x-b)(2(x-a))$$

$$2y\frac{dy}{dx} = 2(x-a)^2 - 2(x-b)(x-a)$$

$$\frac{dy}{dx} = \frac{2(x-a)^2 - 2(x-b)(x-a)}{2y}$$