Implicit Function Theorem second derivative calculation help

Solution 1:

Here is a somewhat more elaborate answer providing two variations. The first is somewhat lengthy since we also write down the arguments. But this way it might be easier to see what is going on. The second one is the more compact notation without arguments which is typically used.

We consider a bivariate function $f(x,y)$ with $y=y(x)$ and \begin{align*} f(x,y(x))=0\tag{1} \end{align*}

We obtain using the chain rule

\begin{align*} \frac{d}{dx}f(x,y(x))&=f_x(x,y(x))+f_y(x,y(x))y^\prime(x)\tag{2}\\ \end{align*}

In short without writing arguments and using (1) we obtain

\begin{align*} \frac{d}{dx}f&=f_x+f_yy^\prime=0\\ y^\prime&=-\frac{f_x}{f_y}\tag{3} \end{align*}

We calculate the second derivative by repeated application of (2).

We obtain \begin{align*} \frac{d^2}{dx^2}f(x,y(x))&=\frac{d}{dx}\left(f_x(x,y(x))+f_y(x,y(x))y^\prime(x)\right)\tag{4}\\ &=\frac{d}{dx}\left(f_x(x,y(x)\right)+\frac{d}{dx}\left(f_y(x,y(x))y^\prime(x)\right)\tag{5}\\ &=f_{xx}(x,y(x))+f_{xy}(x,y(x))y^\prime(x)\\ &\qquad+\left(\frac{d}{dx}f_y(x,y(x))\right)y^\prime(x)+f_y(x,y(x))\frac{d}{dx}\left(y^{\prime}(x)\right)\tag{6}\\ &=f_{xx}(x,y(x))+f_{xy}(x,y(x))y^\prime(x)\\ &\qquad+\Big(f_{yx}(x,y(x))+f_{yy}(x,y(x))y^\prime(x)\Big)y^\prime(x)+f_y(x,y(x))y^{\prime\prime}(x)\tag{7}\\ &=f_{xx}(x,y(x))+\Big(f_{xy}(x,y(x))+f_{yx}(x,y(x))\Big)y^\prime(x)\\ &\qquad+f_{yy}(x,y(x))\left(y^\prime(x)\right)^2 +f_y(x,y(x))y^{\prime\prime}(x)\tag{8}\\ &=0 \end{align*}

Comment:

• In (4) we use (2).

• In (5) we use the linearity of the differential operator.

• In (6) we again apply (2) to the left-hand summand und apply the product rule to the right-hand summand.

• In (7) we again apply (2) and do some rearrangements in (8).

In short and without writing arguments we obtain corresponding to (4) - (8) \begin{align*} \frac{d^2}{dx^2}f&=\frac{d}{dx}\left(f_x+f_yy^\prime\right)\\ &=\frac{d}{dx}f_x+\frac{d}{dx}\left(f_yy^\prime\right)\\ &=f_{xx}+f_{xy}y^\prime+\left(\frac{d}{dx}f_y\right)y^\prime+f_y\left(\frac{d}{dx}y^{\prime}\right)\\ &=f_{xx}+f_{xy}y^\prime+\left(f_{yx}+f_{yy}y^\prime\right)y^\prime+f_yy^{\prime\prime}\\ &=f_{xx}+\left(f_{xy}+f_{yx}\right)y^\prime+f_{yy}\cdot\left(y^\prime\right)^2 +f_yy^{\prime\prime}\\ &=0 \end{align*}

Since $y^\prime=-\frac{f_x}{f_y}$ according to (3) we finally obtain

\begin{align*} y^{\prime\prime}&=-\frac{1}{f_y}\left(f_{xx}+\left(f_{xy}+f_{yx}\right)y^\prime+f_{yy}\cdot\left(y^\prime\right)^2\right)\\ &=-\frac{1}{f_y^3}\left(f_{xx}f_y^2-\left(f_{xy}+f_{yx}\right)f_xf_y+f_{yy}f_x^2\right) \end{align*}

Solution 2:

$y' = -\frac{f_x}{f_y}$ \begin{align} y''&=\frac{\partial y'}{\partial x} + \frac{\partial y'}{\partial y} y' \\ &= \frac {f_x f_{xy} - f_yf_{xx}}{f_y^2}+\frac {f_x f_{yy} - f_yf_{xy}}{f_y^2}y'\\ &= \frac {f_xf_y f_{xy} - f_y^2f_{xx}}{f_y^3}-\frac {f_x^2 f_{yy} - f_xf_yf_{xy}}{f_y^3}\\ &= \frac {2f_xf_y f_{xy} - f_y^2f_{xx}-f_x^2 f_{yy}}{f_y^3} \end{align}