Third Order Implicit Derivative

Question

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!

Answer 1

OK so you have

$y^{''}=-(x^{2}+y^{2})/y^{3}$.

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 :)

Answer 2

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: $$y'=-\frac{x}{y},\; y''=-\frac{x^2+y^2}{y^3},$$ $$y'''=-\frac{3x^3+3xy^2}{y^5}=-\frac{3x(x^2+y^2)}{y^5}=-\frac{27x}{y^5}$$