Second Derivative using Implicit Differentiation

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Homework Help Overview

The problem involves finding the second derivative \( y''(x) \) of the parametric equation \( 9x^2 + y^2 = 9 \) using implicit differentiation. The original poster has already derived the first derivative \( y'(x) = -\frac{9x}{y} \) and is attempting to find the second derivative.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the product rule and substitution in their calculations for \( y''(x) \). Some participants suggest using the quotient rule and looking for clever substitutions to simplify the expression. Others encourage putting results over a common denominator for simplification.

Discussion Status

The discussion has progressed with participants providing tips and suggestions that have helped the original poster move closer to a solution. There is acknowledgment of the complexity of the problem and the need for creative strategies. The original poster expresses gratitude for the feedback received.

Contextual Notes

Participants note the presence of multiple fractions and the potential for simplification, indicating that the problem may involve intricate algebraic manipulation. The original equation's constraints are also referenced in the context of substitutions made during the process.

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Homework Statement


Find y''(x) of the parametric equation 9x^2+y^2=9 using implicit differentiation.

Homework Equations


I already came up with y'(x) = -9x/y

The Attempt at a Solution


Here is what I have for y''(x) so far

y''(x) = d/dx (-9xy^-1)
=-9(d/dx)(xy^-1)
=-9(x(d/dx)(y^-1)+(y^-1)(dx/dx))
=-9(-x(y^-2)y'(x)+y^-1)

I substituted the value of y'(x) = -9x/y here

=-9((-x/(y^2))(-9x/y)+(1/y))
=-9((9x^2)/(y^3)+(1/y))
=(-81x^2)/(y^3)-(9/y)

I know this is incorrect, I originally tried this using the quotient rule but was getting the same answer and the work was much more jumbled, so I opted for the product rule.

The book states the answer is -81/(y^3). I am stuck and haven't been able to work towards the right answer.
 
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Use the quotient rule after you take the first derivative, then look for any possible substitutions that you think might work (& don't be afraid of fractions on fractions on fractions lol).

When you've got it as clean as you can get it (it isn't a messy equation but be careful w/ minus signs) look to see if multiplying the equation (i.e. top & bottom) by a clever choice of 1 will help.

let us know how it goes.
 
You're good so far. Put your result over a common denominator and simplify the top.
 
Ok, I got the answer. The tip on substitution was what really allowed my to break this one open. Since I think this is a great problem that requires a creative strategy (and possibly multiple attempts) to get the simplest answer, I am going to post the rest of my work here.

The result of the Product or Quotient Rule and the substitution for y'(x) yields:

y''(x)=(-9y-((81x^2)/y))/(y^2)

From the original equation, x^2=1-(y^2)/9
I used this value to substitute for x^2.

=(-9y-81((1-(y^2)/9)/(y^2)))/(y^2)

There are a lot of fractions going on, but from here it is just a matter of reducing and simplification.

=(-9y-81(1/y-(y^2)/9y))/(y^2)
=(-9y-81/y+81y/9)/(y^2)
=(-9y-81/y+9y)/(y^2)
=(-81/y)/(y^2)
=-81/y*1/(y^2)
y''(x)=-81/(y^3)

Thanks for the prompt tips. Your feedback helped me solve this problem and find the simplest way to express y''(x).
=
 

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