So, I'm supposed to solve for y''' of the function, $x^2 + y^2 = 9$.

I was able to solve for the second order derivative using implicit differentiation, resulting in:

$y^{''} = (\frac{-y^{2}-x^{2}}{y^{3}})$

Now, I'm a little confused, as I'm not sure if my answer for the third order is correct. To calculate for the third order implicit derivative, will I just use the quotient rule? Doing so, I got:

$y^{'''} = (\frac{{y^{4}{y^{'}}+2xy^{3}}-3x^{2}y^{2}y^{'}}{y^{6}})$

Is this correct? Or can This still be simplified?

Thank you in advance!

  • $\begingroup$ "can This still be simplified?" Sure. Can you see the common factor of $y^2$? $\endgroup$ – Arthur Sep 3 '18 at 6:40
  • $\begingroup$ Oh, I didn't notice that. Thanks! Factoring out $y^{2}$ gives me: $y^{'''} = (\frac{{y^{2}{y^{'}}+2xy}-3x^{2}y^{'}}{y^{4}})$ Will this be the final answer, or will I need to substitute back $y^{'}$ and $y$ back into the equation? $\endgroup$ – Ryan Sep 3 '18 at 6:45
  • $\begingroup$ I would substitute back in for $y^{'}=-x/y$ as well. $\endgroup$ – Simon Terrington Sep 3 '18 at 6:55
  • $\begingroup$ In other news, I also get a different result when I differentiate $y^{''}$ by the quotient rule. It might be worth another look. I get a different numerator. $\endgroup$ – Simon Terrington Sep 3 '18 at 6:59
  • $\begingroup$ Also, I would substitute for $y^{'}$ before you cancel the $y$s. $\endgroup$ – Simon Terrington Sep 3 '18 at 7:01

OK so you have


I agree. But we know that $x^{2}+y^{2}=9$ and so $y^{''}=-9/y^{3}=-9y^{-3}$.

Now another round of implicit differentiation and substituting back in for $y^{'}=-x/y$ gets you there :)

  • $\begingroup$ Now I see! I arrived at the answer: $y^{'''}=\frac{x}{27y^{5}}$ Thank you very much! $\endgroup$ – Ryan Sep 3 '18 at 7:15
  • $\begingroup$ It's a pleasure. It was an interesting problem! $\endgroup$ – Simon Terrington Sep 3 '18 at 18:33

Let $y=y(x)$. Then when differentiate $x^2+y^2=9$ we get $$2yy'+2x=0$$ $$2yy''+2y'^2+2=0$$ $$2yy'''+6y'y''=0$$ Next solve this system: $$y'=-\frac{x}{y},\; y''=-\frac{x^2+y^2}{y^3},$$ $$y'''=-\frac{3x^3+3xy^2}{y^5}=-\frac{3x(x^2+y^2)}{y^5}=-\frac{27x}{y^5}$$


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