finding derivatives of variables in multivariable taylor polynomial

Question

Given: $$F(x,y) = -6 -4(x-4) +6(y-6) +8(x-4)^2 +9(x-4)(y-6) -4(y-6)^2 + R_2$$ and that $F(4,6)=-6$
find y'(4), y''(4), make sure that the function is applicable for the implicit function theorem.
my attempt: so I've basically made sure that the conditions applied, implicitly derived F and found out y'. It was a lot of hard work and while I got a result my method didn't at all utilize the fact it's taylor -I think there must be an easier way but I just can't put my finger on it. I think it'll utilize the fact that $-4 = F_x(x,y)$ and that $6 = F_y(x,y)$ - and maybe the chain rule? since $$-4 = \frac{\partial F(4,6)}{\partial{x}}=\frac{\partial{y}}{\partial{x}}* \frac{\partial F(4,6)}{\partial{y}} = 6*y'(x) \Rightarrow y'(x) = \frac{-2}3$$ but both my calculation and this calculator https://www.emathhelp.net/en/calculators/calculus-1/implicit-differentiation-calculator/?f=-4%28x-4%29%2B6%28y-6%29%2B8%28x-4%29%5E2+%2B9%28x-4%29%28y-6%29+-4%28y-6%29%5E2%3D0&type=x&px=4&py=6 output $2/3$
To sum up - where is my logic faulty? is there a similar easier way? how do I extend this to y''(4)?

Answer 1

From the fact that $$ F(x,y)=F(4,6)+F_x(4,6)(x-4)+F_y(4,6)(y-6)+\frac12F_{xx}(4,6)(x-4)^2 $$ $$ +\frac12F_{yy}(4,6)(y-6)^2+F_{xy}(4,6)(x-4)(y-6)+R_2 $$ it follows that $$ F_x(4,6)=-4,\; F_y(4,6)=6,\; \frac12F_{xx}(4,6)=8,\; F_{xy}(4,6)=9,\; \frac12 F_{yy}(4,6)=-4. $$ Then the derivative of the implicit function is $$ \left.y'(x)\right|_{(4,6)}=-\frac{F_x(4,6)}{F_y(4,6)}=-\frac{-4}{6}=\frac23. $$ Update begin

This formula for the derivative of an implicit function of two variables follows from the multivariable chain rule. $F(x,y(x))\equiv C$ implies that $$ (F(x,y(x)))'=0 $$ or $$ \frac{\partial F}{\partial x}\cdot 1+\frac{\partial F}{\partial y}\cdot y'(x)=0; $$ therefore, $$ y'(x)=-\left. \frac{\partial F}{\partial x} \right/ \frac{\partial F}{\partial y}. $$ Update end

To calculate the second-order derivative, we can use the chain and quotient rules: $$ y''(x)=(y'(x))'=\left(\left.\left( -\frac{F_x}{F_y} \right)\right|_{y=y(x)}\right)'= \left( -\frac{F_x(x,y(x))}{F_y(x,y(x))} \right)' $$ (mostly skip $(x,y(x))$ for brevity) $$ (F_x(x,y(x)))'=F_{xx}\cdot 1+F_{xy}\cdot y'(x) =F_{xx}-F_{xy}\cdot\frac{F_x}{F_y} $$ $$ (F_y(x,y(x)))'=F_{yx}\cdot 1+F_{yy}\cdot y'(x) =F_{yx}-F_{yy}\cdot\frac{F_x}{F_y} $$ Thus, $$ y''(x)=-\frac{(F_x(x,y(x)))'F_y-F_x (F_y(x,y(x)))'}{(F_y)^2} $$ $$ =-\frac{\left(F_{xx}-F_{xy}\cdot\frac{F_x}{F_y}\right)F_y-F_x \left(F_{yx}-F_{yy}\cdot\frac{F_x}{F_y}\right)}{(F_y)^2}. $$ So $$ \left.y''(x)\right|_{(4,6)}=-\frac{\left(16+9\cdot\frac23\right)6+4 \left(9-8\cdot\frac23\right)}{36}. $$