Question #54416

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Answer 1 · 2014-12-08T18:12:15

$y^('')=2$

To solve this problem I will use 3 techniques:

Implicit differentiation. This involves differentiation of both sides of the equation.
The product rule : $(u.v)'=v.(du)/(dx)+u.(dv)/(dx)$
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:

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

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

$y''=2$