How do you find y'' by implicit differentiation for #4x^3 + 3y^3 = 6#?

Question

Emilia Aguilar · Accepted Answer

Please see below.

Starting with: #4x^3+3y^3=6# Differentiate both sides with respect to #x#: #12x^2+9y^2dy/dx=0# Solve for #dy/dx# #" "# (See Note 1 , below) #dy/dx = (-12x^2)/(9y^2)#, so #dy/dx = (-4x^2)/(3y^2)# Differentiate again, using the quotient rule to get #(d^2y)/dx^2 = ((-8x)(3y^2)-(-4x^2)(6y dy/dx))/(3y^2)^2# # = (-24xy^2+24x^2ydy/dx)/(9y^4)# I prefer to remove the common factor before proceeding: # = -24xy((y-x dy/dx)/(9y^4))# Now, replace #dy/dx# # = -24xy((y-x (-4x^2)/(3y^2))/(9y^4))# # = -24xy((y + (4x^3)/(3y^2))/(9y^4))# Now, simplify the complex fraction using your chosen technique. # = -24xy(((y + (4x^3)/(3y^2))(3y^2))/((9y^4)(3y^2)))# # = -24xy((3y^3 + 4x^3)/(27y^6))# I see that I can reduce the fraction, but before I do there's a step I can do to simplify a lot. Way back at the start of the problem, we were told that #4x^3+3y^3=6# So the numerator of our fraction is #6#. #" "# (See Note 2 below.) # = -24xy((6)/(27y^6))# Now simplify the quotient: #(d^2y)/dx^2 = -(16x)/(3y^5)# Note 1 Although we could differentiate again immediately, I prefer not to. If we differentiate without solving for #dy/dx# first, we will need to be careful to disti nguish #(dy/dx)^2# from #(d^2y)/dx^2#. We get #24x +18ydy/dx dy/dx +9y^2 (d^2y)/dx^2 = 0#. It works, but it's kind of a mess. Note 2 This step is typical of certain kinds of implicit differentiation second derivative problems. If you remember to look for it, it can simplify the result considerably.

Emma Aldridge · Answer

undefined To find $ y'' $ by implicit differentiation for the equation $ 4x^3 + 3y^3 = 6 $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $ using the chain rule.
2. Solve for $ \frac{d^2y}{dx^2} $, which represents $ y'' $.