What is the Second Derivative of Implicitly Differentiated Function?

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Homework Help Overview

The problem involves finding the second derivative of a function defined implicitly by the equation 5x² + 3y² = 4. The original poster attempts to differentiate the equation to find y' and subsequently y''.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster first applies the power rule and chain rule to find the first derivative, y', and then attempts to find the second derivative using the quotient rule. They express confusion about how to eliminate variables from their expression for y''.
  • One participant suggests substituting the expression for y' into the equation for y'' to simplify it further.
  • The original poster later describes their process of substitution and simplification, including using the original equation to express x² in terms of y.

Discussion Status

The discussion has progressed with participants exploring substitution and simplification techniques. The original poster has made significant progress in simplifying their expression for y'', but there is no explicit consensus on the final form or correctness of the answer.

Contextual Notes

The original poster expresses uncertainty about how to eliminate x and y' from their final expression for y''. There is an implication of multiple choice answers that do not include these variables, which adds to the complexity of the problem.

Lion214
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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.
 
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Substitute the form of y' you have into the expression for y'' and simplify.
 
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.
 
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.
 

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