Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 60007 by meme last updated on 17/May/19

df of f(x)=x^x y^y

$${df}\:{of}\:{f}\left({x}\right)={x}^{{x}} {y}^{{y}} \\ $$

Answered by alex041103 last updated on 18/May/19

if you mean f(x,y)=x^x y^y  then  df=(∂f/∂x)dx+(∂f/∂y)dy=  =y^y (∂/∂x)(x^x )dx+x^x (∂/∂y)(y^y )dy=  In general let′s find (∂/∂z)z^z :  z^z =e^(z ln(z)) ⇒(∂/∂z)(z^z )=(∂/∂z)(e^(z ln(z)) )=  =(∂/(∂(z ln(z))))(e^(z ln(z)) )×((∂(z ln(z)))/∂z)=  =e^(z ln(z)) ×(1×ln(z)+z×(1/z))=  =z^z (1+ln(z))=((∂(z^z ))/∂z)  ⇒df=x^x y^y (1+ln(x))dx+x^x y^y (1+ln(y))dy  ⇒df=f[(1+ln(x))dx+(1+ln(y))dy]  you can go a little bit further:  (df/f)=dx+ln(x)dx+dy+ln(y)dy=  =d(x+y)+ln(x^dx )+ln(y^dy )=  =d(x+y)+ln(x^dx y^dy )  ⇒df=x^x y^y [d(x+y)+ln(x^dx y^dy )]

$${if}\:{you}\:{mean}\:{f}\left({x},{y}\right)={x}^{{x}} {y}^{{y}} \:{then} \\ $$$${df}=\frac{\partial{f}}{\partial{x}}{dx}+\frac{\partial{f}}{\partial{y}}{dy}= \\ $$$$={y}^{{y}} \frac{\partial}{\partial{x}}\left({x}^{{x}} \right){dx}+{x}^{{x}} \frac{\partial}{\partial{y}}\left({y}^{{y}} \right){dy}= \\ $$$${In}\:{general}\:{let}'{s}\:{find}\:\frac{\partial}{\partial{z}}{z}^{{z}} : \\ $$$${z}^{{z}} ={e}^{{z}\:{ln}\left({z}\right)} \Rightarrow\frac{\partial}{\partial{z}}\left({z}^{{z}} \right)=\frac{\partial}{\partial{z}}\left({e}^{{z}\:{ln}\left({z}\right)} \right)= \\ $$$$=\frac{\partial}{\partial\left({z}\:{ln}\left({z}\right)\right)}\left({e}^{{z}\:{ln}\left({z}\right)} \right)×\frac{\partial\left({z}\:{ln}\left({z}\right)\right)}{\partial{z}}= \\ $$$$={e}^{{z}\:{ln}\left({z}\right)} ×\left(\mathrm{1}×{ln}\left({z}\right)+{z}×\frac{\mathrm{1}}{{z}}\right)= \\ $$$$={z}^{{z}} \left(\mathrm{1}+{ln}\left({z}\right)\right)=\frac{\partial\left({z}^{{z}} \right)}{\partial{z}} \\ $$$$\Rightarrow{df}={x}^{{x}} {y}^{{y}} \left(\mathrm{1}+{ln}\left({x}\right)\right){dx}+{x}^{{x}} {y}^{{y}} \left(\mathrm{1}+{ln}\left({y}\right)\right){dy} \\ $$$$\Rightarrow{df}={f}\left[\left(\mathrm{1}+{ln}\left({x}\right)\right){dx}+\left(\mathrm{1}+{ln}\left({y}\right)\right){dy}\right] \\ $$$${you}\:{can}\:{go}\:{a}\:{little}\:{bit}\:{further}: \\ $$$$\frac{{df}}{{f}}={dx}+{ln}\left({x}\right){dx}+{dy}+{ln}\left({y}\right){dy}= \\ $$$$={d}\left({x}+{y}\right)+{ln}\left({x}^{{dx}} \right)+{ln}\left({y}^{{dy}} \right)= \\ $$$$={d}\left({x}+{y}\right)+{ln}\left({x}^{{dx}} {y}^{{dy}} \right) \\ $$$$\Rightarrow{df}={x}^{{x}} {y}^{{y}} \left[{d}\left({x}+{y}\right)+{ln}\left({x}^{{dx}} {y}^{{dy}} \right)\right] \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com