Question #54416

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 : $\left(u . v\right) ' = v . \frac{\mathrm{du}}{\mathrm{dx}} + u . \frac{\mathrm{dv}}{\mathrm{dx}}$

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:

$D \left({x}^{2} + x y + {y}^{3}\right) = D \left(1\right)$

I'll differentiate each term:

$\left({x}^{2}\right) ' = 2 x$

$\left(x y\right) ' = x \frac{\mathrm{dy}}{\mathrm{dx}} + y \frac{\mathrm{dx}}{\mathrm{dx}} = x y ' + y$ (Product rule)

$\left({y}^{3}\right) ' = 3 {y}^{2.} y '$ (chain rule)

$\left(1\right) ' = 0$

So:
$2 x + x y ' + y + 3 {y}^{2} y ' = 0$

Factorising we get:

$y ' \left(x + 3 {y}^{2}\right) = - \left(2 x + y\right)$
Eqn 1.

Now we differentiate again using the same techniques:

$y ' ' \left(x + 3 {y}^{2}\right) + y ' \left(1 + 6 y y '\right) = - \left(2 + y '\right)$

This simplifies to:

$y ' ' \left(x + 3 {y}^{2}\right) = - y ' - 6 y {\left(y '\right)}^{2} - 2 - y '$

So:
$y ' ' \left(x + 3 {y}^{2}\right) = - y ' \left(2 + 6 y y '\right) - 2$
Eqn 2.

The original equation is ${x}^{2} + x y + {y}^{3} = 1$

So if $x = 1$ $y = 0$

We can put these values back into Eqn 1:

$y ' \left(1 + 3 {\left(0\right)}^{2}\right) = - \left(2 \left(1\right) + 0\right)$

Hence $y ' = - 2$

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

So:

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

$y ' ' = 2 \left(2 + 0\right) - 2$

$y ' ' = 2$