I'm trying to compute the implicit function theorem's second derivative but I'm getting stuck.

We have a function $f(x, y)$ where $y(x)$ and we know that $$\frac{dy}{dx}=-\frac{f_x}{f_y}$$.

And I'm trying to get to $y''$ which according to the book is $$y'' = \frac{-f_y^2f_{xx}+ 2f_xf_yf_{xy}-f_x^2f_{yy}}{f_y^3}$$

But I'm somehow messing up the partial derivatives:

$$y'f_y + f_x = 0$$ $$y''f_y + y'f_{yx} + f_{xx}y' $$ ??

I forgot how to properly go about computing these partial derivatives


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*}


  • 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*}


$y' = -\frac{f_x}{f_y}$ \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}


i have $$f_{xx}+f_{xy}y'+f_{yy}y'^2+f_{y}y''+f_{yx}y'=0$$ from here we get $$y''=\frac{f_xf_y(f_{xy}+f_{yx})-f_{xx}f_y^2-f_{yy}f_x^2}{f_y^3}$$

  • $\begingroup$ Can you explain how you are computing the partial derivatives? I forgot how to go about that. $f_x' = f_{xx}$ right? How do you get the other four terms from? $\endgroup$ – user366746 Nov 30 '16 at 19:07

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