Question

Question #54416

Answer 1

`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`