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The derivative of the equation $x^2-y^3=4y+9$ can be found using implicit differentiation to yield $y'=2x/(4+3y^2)$. (I think.) I am tasked with showing where the original function is rising, falling, and resting using the equation for its derivative as shown above.

The answer I found in this forum uses tools that stray far from the contents of my toolbox. I did devise an answer on my own: decreasing $(-\infty$, -3), increasing $(-3,3)$, and falling again $(3,\infty)$. Even if that answer is correct, however, I found it through some rather unseemly jury-rigging that did not involve the derivative.

I suspect a more elegant solution exists.

N.B. I come to this problem by dint of working through Paul's Online Notes on my own, just for yucks and giggles. It can be found here: PON Assignment Problem

Omne Bonum
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    What is the problem in concluding that for negative value of x it is decreasing and vice versa? – xrfxlp Jun 20 '19 at 15:59
  • To check your results you can solve $$y^3+4y+9-x^2=0$$ for $y$ – Dr. Sonnhard Graubner Jun 20 '19 at 16:05
  • Your question is to find when the original function is rising, falling, or resting, but you appear to be attempting to find when the derivative is rising, falling, or decreasing. You would calculate this using the second derivative: $$y'' = \dfrac{2(4+3y^2)^2-24x^2y}{(4+3y^2)^3}$$ Set that equal to zero. Using the fact that $x^2=y^3+4y+9$ would give you a 4th degree polynomial in $y$. Solving that would give you the $y-$ coordinates when concavity changes. You are correct that it is very close to $x=-3$ and $x=3$. But, that is not really what the problem is asking. – SlipEternal Jun 20 '19 at 16:46
  • @InterstellarProbe To clarify, I'm only interested in finding where the first derivative of the function is positive or negative, which would indicate where the original function is rising or falling, respectively. Sorry for the ambiguity. – Omne Bonum Jun 20 '19 at 17:58
  • @OmneBonum I understand what the problem states. My explanation was to try to help you understand where your attempts were leading you because you were on the right track to an answer to a different question. – SlipEternal Jun 20 '19 at 18:03
  • @Dr.SonnhardGraubner The task is to use implicit differentiation first to answer the question about the up and down behavior of the original function. How do I use $y'=2x/(4+3y^2)$ to find the answer before solving for y to check my answer? – Omne Bonum Jun 20 '19 at 18:04
  • @Omnebrown, you are looking for $y'>0$ and $y'<0$. The denominator is always greater than zero. $4+3y^2$ is a positive number plus a nonnegative number ($3y^2\ge 0$ for all $y$). So, $y'$ will be positive when the numerator is positive and negative when the numerator is negative. As Ajay Mishra states in the answer below, $f(x)$ decreases on $(-\infty,0)$ and $f(x)$ increases on $(0,\infty)$. – SlipEternal Jun 20 '19 at 18:07
  • That is the elegant solution I sought. I had no idea it would be so simple! Now I feel silly. Cheers. – Omne Bonum Jun 20 '19 at 18:21

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enter image description hereAs you have calculated $$ y’ = \cfrac{2x}{4+3y^2}$$ In which only variable which changes the sign of derivative is x, as $y^2$ is always positive, so it is same to conclude that when $x<0,y$ is decreasing and vice versa.

xrfxlp
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