# Implicit Differentiation.

1. Oct 5, 2013

### Lion214

1. The problem statement, all variables and given/known data

Determine y'' when 5x^2 + 3y^2 = 4.

3. The attempt at a solution

So I found the first derivative using the power rule and chain rule,

10x + 6yy' = 0

Which I then solved for y',

y' = -10x/6y = -5x/3y

Next I found the second derivative using quotient rule,

y'' = (-15y + 15xy')/(3y)^2

This is the part where I am lost, since all the multiple choice answers involve no y' nor is there a x.
I don't know how to get rid of the x and the y' in the equation. Any help will be appreciated as to simplifying the equation even further and as to how.

2. Oct 5, 2013

Substitute the form of y' you have into the expression for y'' and simplify.

3. Oct 5, 2013

### Lion214

so here's what I did,

I substituted y' with -5x/3y and got

y'' = (-15y + 15x(-5x/3y))/(3y)^2 = (-15y - 25x^2/y)/(3y)^2 = (-15y^2/y - 25x^2/y)/(9y^2) = (-15y^2-25x^2)/9y^3

But I still had to get rid of the x, so I used the original equation to solve for x^2, which was;

x^2 = (4-3y^2)/5

Which I then use to simplify even further,

y'' = (-15y^2 - 25((4 - 3y^2)/5))/9y^3 = ((-75y^2 - 100 + 75y)/5)/9y^3

In which the answer is -20/9y^3.

Thanks for the insight. For some reason I didn't see that.

4. Oct 5, 2013

You are welcome. The little bit of extra'' work you've just gone through is handy to remember for these types of problems - I would state with near 100% certainty you'll see similar things in the future.