0
$\begingroup$

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 $f$

$\endgroup$

2 Answers 2

2
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.