# Second derivative of multivariable implicit function

Find $$d^2z$$ of the following function $$\frac{x}{z}= \ln{\frac{z}{y}}+1$$

Finding the total derivative of the left and right side we arrive at the expression:

$$\frac{zdx-xdz}{z^2}=\frac{y}{z} \cdot \frac{ydz-zdy}{y^2} \iff yzdx-xydz-yzdz+z^2dy=0$$

Differentiating again this expression:

$$y(x+z) d^2z=zdzdy+(zdy-xdy)dz - y^2dz^2$$

Also $$dz=\frac{z(ydx+zdy)}{y(x+z)}$$ so by substituting in the last equation we get:

$$d^2z= - \frac{z^2(ydx-xdy)^2}{y^2(x+z)^3}$$

I have couple of questions regarding this solution. I've managed to replicate taking the first derivative from both sides of the equation and arrived at the same result. What I noticed, is that during this we do not consider $$z$$ as a function of $$x$$ and $$y$$.

Next when taking the second derivative I didn't get the extra term $$y(x+z)d^2z = yx d^2z +z d^2z$$. This term seems to appear by differentiating all the $$dz$$ parts in $$yzdx-xydz-yzdz+z^2dy=0$$. Perhaps it is from somewhere else that we get the $$d^2z$$ term but I don't know what I'm missing.

I think that the best would be to solve for $$z$$ the equation $$\frac{x}{z}= \ln{\frac{z}{y}}+1$$ Its solution is given in terms of Lambert function $$z=\frac{x}{W\left(\frac{e x}{y}\right)}$$ from which $$z'_x=\frac{1}{W\left(\frac{e x}{y}\right)+1}$$ $$z'_y=\frac{x}{y \left(W\left(\frac{e x}{y}\right)^2+W\left(\frac{e x}{y}\right)\right)}$$ $$z''_{xx}=-\frac{W\left(\frac{e x}{y}\right)}{x \left(W\left(\frac{e x}{y}\right)+1\right)^3}$$ $$z''_{xy}=\frac{W\left(\frac{e x}{y}\right)}{y \left(W\left(\frac{e x}{y}\right)+1\right)^3}$$ $$z''_{yy}=-\frac{x W\left(\frac{e x}{y}\right)}{y^2 \left(W\left(\frac{e x}{y}\right)+1\right)^3}$$
• @DreaDk. You will learn it soon ! It is just fascinating. Notice that, if you replace $W(.)$ by $\frac x z$ all derivatives become simple. Cheers :-) May 6, 2019 at 9:40