Second Derivative using Implicit Differentiation

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The discussion focuses on finding the second derivative y''(x) of the parametric equation 9x^2 + y^2 = 9 using implicit differentiation. The initial attempt yielded an incorrect expression for y''(x), prompting the user to seek guidance on applying the quotient rule and making effective substitutions. After receiving feedback, the user successfully simplified their work and derived the correct result for y''(x) as -81/(y^3). The conversation highlights the importance of strategic problem-solving and the value of peer assistance in mathematics. Ultimately, the user expresses gratitude for the tips that led to the correct solution.
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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).
=
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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