# The Chain rule - prove identity

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 $$f$$

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 $$g: \mathbb{R}^2 \to \mathbb{R}$$, $$f: \mathbb{R} \to \mathbb{R}$$ $$\frac{\partial}{\partial x} f(g(x,y)) = \frac{\partial g}{\partial x} f'(g(x,y))$$

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

• But in what I was asked to prove f doesn't appear Apr 13 at 10:25
• 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. Apr 13 at 10:36