What is the Second Derivative of Implicitly Differentiated Function?

[Lion214]

Homework Statement

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

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.

 

[statdad]

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

 

[Lion214]

Thanks statdad!

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.

 

[statdad]

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.