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Question Number 74663 by TawaTawa last updated on 28/Nov/19
If x^x y^y z^z = c show that at x = y = z (∂^2 z/(∂x∂y)) = − (x log ex)^(−1)
$$\mathrm{If}\:\:\:\:\mathrm{x}^{\mathrm{x}} \:\mathrm{y}^{\mathrm{y}} \:\mathrm{z}^{\mathrm{z}} \:\:\:=\:\:\:\mathrm{c}\:\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\:\mathrm{at}\:\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{y}\:\:=\:\:\mathrm{z} \\ $$$$\:\:\:\:\:\:\frac{\partial^{\mathrm{2}} \mathrm{z}}{\partial\mathrm{x}\partial\mathrm{y}}\:\:\:=\:\:\:−\:\left(\mathrm{x}\:\mathrm{log}\:\mathrm{ex}\right)^{−\mathrm{1}} \\ $$
Answered by mind is power last updated on 28/Nov/19
⇒zlog(z)=log(c)−xlog(x)−ylog(y) let f(z)=zlog(z)=log(c)−xlog(x)−ylog(y) ⇒log(z)e^(log(z)) =f(z) ⇒log(z)=W(f(z))⇒z=e^(W(f)) ,W lambertfunction W′(t)=((W(t))/(t(1+W(t)))) Z=e^(W(f(x,y))) ⇒W(f)=log(z) (∂z/∂y)=(∂f/∂y)W′(f(x,y))e^(W(f)) =(−log(y)−1)((We^(W(f)) )/(f(1+W(f)))) =((−(log(y)+1).)/((1+W(f)))) (∂^2 Z/(∂x∂y))=(∂z/∂x).((−(log(y)+1))/(1+W(f)))=(log(y)+1).(((∂f/∂x).W′(f))/((1+W(f))^2 )) =(log(y)+1).(((−1−log(x)).W(f))/(f(x,y).{1+W(f)}^3 )) =((−(1+log(y))(1+log(x)).W(f(x,y)))/(zlogz(1+log(z))^3 )). =((Log(z))/(zlogz(1+log(x))) =−(z+zlog(e)(z))^(−1) =−(x+xln(x))^(−1) x(1+ln(x))=x(ln(xe)⇒ =−(xln(xe))^(−1)
$$\Rightarrow\mathrm{zlog}\left(\mathrm{z}\right)=\mathrm{log}\left(\mathrm{c}\right)−\mathrm{xlog}\left(\mathrm{x}\right)−\mathrm{ylog}\left(\mathrm{y}\right) \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{zlog}\left(\mathrm{z}\right)=\mathrm{log}\left(\mathrm{c}\right)−\mathrm{xlog}\left(\mathrm{x}\right)−\mathrm{ylog}\left(\mathrm{y}\right) \\ $$$$\Rightarrow\mathrm{log}\left(\mathrm{z}\right)\mathrm{e}^{\mathrm{log}\left(\mathrm{z}\right)} =\mathrm{f}\left(\mathrm{z}\right) \\ $$$$\Rightarrow\mathrm{log}\left(\mathrm{z}\right)=\mathrm{W}\left(\mathrm{f}\left(\mathrm{z}\right)\right)\Rightarrow\mathrm{z}=\mathrm{e}^{\mathrm{W}\left(\mathrm{f}\right)} ,\mathrm{W}\:\mathrm{lambertfunction} \\ $$$$\mathrm{W}'\left(\mathrm{t}\right)=\frac{\mathrm{W}\left(\mathrm{t}\right)}{\mathrm{t}\left(\mathrm{1}+\mathrm{W}\left(\mathrm{t}\right)\right)}\: \\ $$$$\mathrm{Z}=\mathrm{e}^{\mathrm{W}\left(\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\right)} \\ $$$$\Rightarrow\mathrm{W}\left(\mathrm{f}\right)=\mathrm{log}\left(\mathrm{z}\right) \\ $$$$\frac{\partial\mathrm{z}}{\partial\mathrm{y}}=\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\mathrm{W}'\left(\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\right)\mathrm{e}^{\mathrm{W}\left(\mathrm{f}\right)} \\ $$$$=\left(−\mathrm{log}\left(\mathrm{y}\right)−\mathrm{1}\right)\frac{\mathrm{We}^{\mathrm{W}\left(\mathrm{f}\right)} }{\mathrm{f}\left(\mathrm{1}+\mathrm{W}\left(\mathrm{f}\right)\right)} \\ $$$$=\frac{−\left(\mathrm{log}\left(\mathrm{y}\right)+\mathrm{1}\right).}{\left(\mathrm{1}+\mathrm{W}\left(\mathrm{f}\right)\right)} \\ $$$$\frac{\partial^{\mathrm{2}} \mathrm{Z}}{\partial\mathrm{x}\partial\mathrm{y}}=\frac{\partial\mathrm{z}}{\partial\mathrm{x}}.\frac{−\left(\mathrm{log}\left(\mathrm{y}\right)+\mathrm{1}\right)}{\mathrm{1}+\mathrm{W}\left(\mathrm{f}\right)}=\left(\mathrm{log}\left(\mathrm{y}\right)+\mathrm{1}\right).\frac{\frac{\partial\mathrm{f}}{\partial\mathrm{x}}.\mathrm{W}'\left(\mathrm{f}\right)}{\left(\mathrm{1}+\mathrm{W}\left(\mathrm{f}\right)\right)^{\mathrm{2}} } \\ $$$$=\left(\mathrm{log}\left(\mathrm{y}\right)+\mathrm{1}\right).\frac{\left(−\mathrm{1}−\mathrm{log}\left(\mathrm{x}\right)\right).\mathrm{W}\left(\mathrm{f}\right)}{\mathrm{f}\left(\mathrm{x},\mathrm{y}\right).\left\{\mathrm{1}+\mathrm{W}\left(\mathrm{f}\right)\right\}^{\mathrm{3}} } \\ $$$$=\frac{−\left(\mathrm{1}+\mathrm{log}\left(\mathrm{y}\right)\right)\left(\mathrm{1}+\mathrm{log}\left(\mathrm{x}\right)\right).\mathrm{W}\left(\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\right)}{\mathrm{zlogz}\left(\mathrm{1}+\mathrm{log}\left(\mathrm{z}\right)\right)^{\mathrm{3}} }. \\ $$$$=\frac{\mathrm{Log}\left(\mathrm{z}\right)}{\mathrm{zlogz}\left(\mathrm{1}+\mathrm{log}\left(\mathrm{x}\right)\right.} \\ $$$$=−\left(\mathrm{z}+\mathrm{zlog}\left(\mathrm{e}\right)\left(\mathrm{z}\right)\right)^{−\mathrm{1}} =−\left(\mathrm{x}+\mathrm{xln}\left(\mathrm{x}\right)\right)^{−\mathrm{1}} \\ $$$$\mathrm{x}\left(\mathrm{1}+\mathrm{ln}\left(\mathrm{x}\right)\right)=\mathrm{x}\left(\mathrm{ln}\left(\mathrm{xe}\right)\Rightarrow\right. \\ $$$$=−\left(\mathrm{xln}\left(\mathrm{xe}\right)\right)^{−\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by TawaTawa last updated on 28/Nov/19
Wow, God bless you sir.
$$\mathrm{Wow},\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mind is power last updated on 28/Nov/19
y′re welcom
$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$

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