let $z(x,y)=\frac{\ln x}{y^{2}}f\left(\frac{e^{-y}}{x}\right)$ for some differentiable $f(t)$

prove: $$(xy\ln x)\cdot\frac{\partial z}{\partial x}-(y \ln x)\cdot\frac{\partial z}{\partial y}=(2\ln x+y)\cdot z$$

my try:

my idea was to use the Chain rule

let $u=\frac{lnx}{y^{2}},v=\frac{e^{-y}}{x}$ the composition is:

$$z=uf\left(v\right)\longleftarrow(u,v)=(\frac{\ln x}{y^{2}},\frac{e^{-y}}{x})\longleftarrow(x,y)$$

and by the Chain rule:

$$ \begin{pmatrix}\frac{\partial z}{\partial x} & \frac{\partial z}{\partial y}\end{pmatrix}=\begin{pmatrix}\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v}\end{pmatrix}\begin{pmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}$$

but I don't know how to continue because $\frac{\partial z}{\partial v}$ depends on the function $f$


2 Answers 2


Since $z(x,y)=\frac{\ln x}{y^2}f\left(\frac{e^{-y}}x\right)$, you have$$\frac{\partial z}{\partial x}=\frac1{xy^2}f\left(\frac{e^{-y}}x\right)-\frac{e^{-y}\ln x}{x^2y^2}f'\left(\frac{e^{-y}}x\right)$$and$$\frac{\partial z}{\partial y}=-\frac{2\ln x}{y^3}f\left(\frac{e^{-y}}x\right)-\frac{e^{-y}\ln x}{xy^2}f'\left(\frac{e^{-y}}x\right).$$Therefore,$$xy\ln(x)\frac{\partial z}{\partial x}=\frac{\ln x}yf\left(\frac{e^{-y}}x\right)-\frac{e^{-y}\ln^2x}{xy}f'\left(\frac{e^{-y}}x\right)$$and$$y\ln(x)\frac{\partial z}{\partial y}=-\frac{2\ln^2x}{y^2}f\left(\frac{e^{-y}}x\right)-\frac{e^{-y}\ln^2x}{xy}f'\left(\frac{e^{-y}}x\right).$$So\begin{align}xy\ln(x)\frac{\partial z}{\partial x}-y\ln(x)\frac{\partial z}{\partial y}&=\left(\frac{\ln x}y+\frac{2\ln^2x}{y^2}\right)f\left(\frac{e^{-y}}x\right)\\&=\frac{y\ln(x)+2\ln^2(x)}{y^2}f\left(\frac{e^{-y}}x\right)\\&=(y+2\ln x)z(x,y).\end{align}


The final result will depend on $f$ (and $f'$). For instance,

$$ \frac{\partial z}{\partial x} = \frac{1}{y^2}\left[\frac 1x f\left(\frac{e^{-y}}{x}\right) + \ln x \cdot \left(-\frac{e^{-y}}{x^2}\right)\cdot f'\left(\frac{e^{-y}}{x}\right) \right]. $$

In a simpler case, assuming enough regularity and $g: \mathbb{R}^2 \to \mathbb{R}$, $f: \mathbb{R} \to \mathbb{R}$ $$ \frac{\partial}{\partial x} f(g(x,y)) = \frac{\partial g}{\partial x} f'(g(x,y)) $$

$$ \frac{\partial}{\partial y} f(g(x,y)) = \frac{\partial g}{\partial y} f'(g(x,y)) $$

  • $\begingroup$ But in what I was asked to prove f doesn't appear $\endgroup$
    – DanielG
    Apr 13 at 10:25
  • $\begingroup$ Substitute $z$ and its derivatives in the LHS. You'll get an expression involving $f$ and $f'$ and you'll see that $f'$ disappears. After that, recognise what part of the expºression corresponds to $z$ and you will get the RHS. $\endgroup$ Apr 13 at 10:36

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