Question #54416

1 Answer
Dec 8, 2014

#y^('')=2#

To solve this problem I will use 3 techniques:

  1. Implicit differentiation. This involves differentiation of both sides of the equation.

  2. The product rule : #(u.v)'=v.(du)/(dx)+u.(dv)/(dx)#

  3. The chain rule. This involves differentiation of the outer layer and multiplying by the derivative of the inner layer.

I will show how this applies to the 1st derivative then work through the 2nd.

Using rule 1 we can write:

#D(x^2 +xy +y^3)=D(1)#

I'll differentiate each term:

#(x^2)'=2x#

#(xy)'=x(dy)/(dx)+y(dx)/(dx)=xy'+y# (Product rule)

#(y^3)'=3y^2.y'# (chain rule)

#(1)'=0#

So:
#2x+xy'+ y+3y^2y'=0#

Factorising we get:

#y'(x+3y^2)=-(2x+y)#
Eqn 1.

Now we differentiate again using the same techniques:

#y''(x+3y^2)+y'(1+6yy')=-(2+y')#

This simplifies to:

#y''(x+3y^2)=-y'-6y(y')^2-2-y'#

So:
#y''(x+3y^2)=-y'(2+6yy')-2#
Eqn 2.

The original equation is #x^2+xy+y^3=1#

So if #x=1# #y=0#

We can put these values back into Eqn 1:

#y'(1+3(0)^2)=-(2(1)+0)#

Hence #y'=-2#

Since #x=1# and #y=0# we can put these 3 values back into. eqn 2:

So:

#y''(1+0)=-(2)(2+6(0))-2#

#y''=2(2+0)-2#

#y''=2#