Question

If ${\displaystyle z=\log (e^x+e^y)\text{show that rt}-s^2=0 \text{where r}=\frac{{\partial }^{2}z}{\partial x^2},t=\frac{{\partial }^{2}z}{\partial y^2}\text{s}=\frac{{\partial }^{2}z}{\partial x\partial y}}$

Answer 1

${\displaystyle z=\log (e^x+e^y)}$

(1)${\displaystyle \frac{\partial z}{\partial x}=\frac{e^x}{e^x+e^y}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}=\frac{e^x(e^x+e^y)-e^x\left(e^x\right)}{{(e^x+e^y)}^2}=\frac{e^{2x}+e^{xy}-e^{2x}}{{(e^x+e^y)}^2}}$

${\displaystyle r=\frac{{\partial }^{2}z}{\partial x^2}=\frac{e^{xy}}{{(e^x+e^y)}^2}}$  .............(1)

(2) ${\displaystyle \frac{\partial z}{\partial y}=\frac{e^y}{e^x+e^y}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=\frac{e^y(e^x+e^y)-e^y\left(e^y\right)}{{(e^x+e^y)}^2}=\frac{e^{2y}+e^{xy}-e^{2y}}{{(e^2+e^y)}^2}}$

t ${\displaystyle =\frac{{\partial }^{2}z}{\partial y^2}=\frac{e^{xy}}{e^x+e^y}2}$   ................(2)

Answer 2

${\displaystyle z=\log (e^x+e^y)}$

(1)${\displaystyle \frac{\partial z}{\partial x}=\frac{e^x}{e^x+e^y}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}=\frac{e^x(e^x+e^y)-e^x\left(e^x\right)}{{(e^x+e^y)}^2}=\frac{e^{2x}+e^{xy}-e^{2x}}{{(e^x+e^y)}^2}}$

${\displaystyle r=\frac{{\partial }^{2}z}{\partial x^2}=\frac{e^{xy}}{{(e^x+e^y)}^2}}$  .............(1)

(2) ${\displaystyle \frac{\partial z}{\partial y}=\frac{e^y}{e^x+e^y}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=\frac{e^y(e^x+e^y)-e^y\left(e^y\right)}{{(e^x+e^y)}^2}=\frac{e^{2y}+e^{xy}-e^{2y}}{{(e^2+e^y)}^2}}$

t ${\displaystyle =\frac{{\partial }^{2}z}{\partial y^2}=\frac{e^{xy}}{e^x+e^y}2}$   ................(2)

(3) ${\displaystyle \frac{\partial z}{\partial x}=\frac{e^x}{{(e^x+e^y)}^{}}}$

s ${\displaystyle =\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{e^{xy}}{{(e^x+e^y)}^2}}$ ................(3)

From (1), (2) and (3) we get,

rt ${\displaystyle =\left(\frac{e^{xy}}{{(e^x+e^y)}^2}\right)\times \left(\frac{e^{xy}}{{(e^x+e^y)}^2}\right)={\left(\frac{e^{xy}}{{(e^x+e^y)}^2}\right)}^2=\left(\frac{e^{2xy}}{{(e^x+e^y)}^2}\right)}$  .............(4)

s² = ${\displaystyle {\left(\frac{e^{xy}}{{(e^x+e^y)}^2}\right)}^2=\left(\frac{e^{2xy}}{{(e^x+e^y)}^2}\right)}$................(5)

From (4) and (5) we get,

rt-s² = ${\displaystyle \left(\frac{e^{2xy}}{{(e^x+e^y)}^2}\right)-\left(\frac{e^{2xy}}{{(e^x+e^y)}^2}\right)=0}$

Hence proved rt - ${\displaystyle s^2}$= 0

Answer 3

${\displaystyle z=\log (e^x+e^y)}$

(1)${\displaystyle \frac{\partial z}{\partial x}=\frac{e^x}{e^x+e^y}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}=\frac{e^x(e^x+e^y)-e^x\left(e^x\right)}{{(e^x+e^y)}^2}=\frac{e^{2x}+e^{xy}-e^{2x}}{{(e^x+e^y)}^2}}$

${\displaystyle r=\frac{{\partial }^{2}z}{\partial x^2}=\frac{e^{xy}}{{(e^x+e^y)}^2}}$  .............(1)

(2) ${\displaystyle \frac{\partial z}{\partial y}=\frac{e^y}{e^x+e^y}}$

${\displaystyle \frac{{\partial }^{2}z}{\partial y^2}=\frac{e^y(e^x+e^y)-e^y\left(e^y\right)}{{(e^x+e^y)}^2}=\frac{e^{2y}+e^{xy}-e^{2y}}{{(e^2+e^y)}^2}}$

t ${\displaystyle =\frac{{\partial }^{2}z}{\partial y^2}=\frac{e^{xy}}{e^x+e^y}2}$   ................(2)

(3) ${\displaystyle \frac{\partial z}{\partial x}=\frac{e^x}{{(e^x+e^y)}^{}}}$

s ${\displaystyle =\frac{{\partial }^{2}z}{\partial x\partial y}=\frac{e^{xy}}{{(e^x+e^y)}^2}}$ ................(3)

From (1), (2) and (3) we get,

rt ${\displaystyle =\left(\frac{e^{xy}}{{(e^x+e^y)}^2}\right)\times \left(\frac{e^{xy}}{{(e^x+e^y)}^2}\right)={\left(\frac{e^{xy}}{{(e^x+e^y)}^2}\right)}^2=\left(\frac{e^{2xy}}{{(e^x+e^y)}^2}\right)}$  .............(4)

s² = ${\displaystyle {\left(\frac{e^{xy}}{{(e^x+e^y)}^2}\right)}^2=\left(\frac{e^{2xy}}{{(e^x+e^y)}^2}\right)}$................(5)

From (4) and (5) we get,

rt-s² = ${\displaystyle \left(\frac{e^{2xy}}{{(e^x+e^y)}^2}\right)-\left(\frac{e^{2xy}}{{(e^x+e^y)}^2}\right)=0}$

Hence proved rt - ${\displaystyle s^2}$= 0