How do you twice differentiate #x^3+y^3=1#?

Question

9413 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-30T22:26:36

#y''=-(2x)/(y^5)#

#x^(3)+y^(3)=1#

#D[x^(3)+y^(3)]=D[1]#

So:

#3x^2+3y^2y'=0#

#y'=-(cancel(3)x^2)/(cancel(3)y^2)=-(x^2)/(y^(2)# #color(red)((1))#

So now we have to differentiate again using the quotient rule. A useful tip is to start and finish with the function on the bottom #rArr#

#y''=(y^(2)(-2x)-(-x^(2)).2y.y')/(y^(4))# #color(red)((2))#

We already know from #color(red)((1))# that :

#y'=-x^(2)/y^2#

So we can substitute that expression for #y'# into #color(red)((2))# #rArr#

#y''=(y^(2)(-2x)-(-x^(2)).2y.((-x^(2))/(y^(2))))/(y^(4)#

#y''=[-2xy^2-(2x^4)/(y)]/[y^4]#

Multiplying top and bottom by #y# #rArr#

#y''=([-2xy^3-2x^4])/(y^5)#

#y''=(-2x[y^(3)+x^(3)])/(y^5)# #color(red)((3))#

Since we know that:

#x^3+y^3=1#

We can substitute that value of #1# into #color(red)((3))rArr#

#y''=(-2x)/(y^5)#