If x^x y^y z^z = c show that at x = y = z (∂^2 z/(∂x∂y)) = − (x log ex)^(−1)


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Question Number 74663 by TawaTawa last updated on 28/Nov/19

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

	$Wow, God bless you sir .$
	Commented by mind is power last updated on 28/Nov/19

	$y^{'} re welcom$
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