Exercise
Given the differentiable function
z(x,y)=\ln(e^x+e^y)
Prove the equation
z''_{xx}\cdot z''_{yy}={(z''_{xy})}^2
Proof
We will use the chain rule to calculate the partial derivatives of z.
z'_x=\frac{1}{e^x+e^y}\cdot e^x=
=\frac{e^x}{e^x+e^y}
z'_y=\frac{1}{e^x+e^y}\cdot e^y=
=\frac{e^y}{e^x+e^y}
We will calculate the second order derivatives.
z''_{xx}=\frac{e^x(e^x+e^y)-e^x\cdot e^x}{{(e^x+e^y)}^2}=
=\frac{e^x\cdot e^y}{{(e^x+e^y)}^2}
z''_{yy}=\frac{e^y(e^x+e^y)-e^y\cdot e^y}{{(e^x+e^y)}^2}=
=\frac{e^y\cdot e^x}{{(e^x+e^y)}^2}
z''_{xy}=\frac{-e^x\cdot e^y}{{(e^x+e^y)}^2}
We will put the partial derivatives in the left side of the equation we need to prove.
z''_{xx}\cdot z''_{yy}=
=\frac{e^x\cdot e^y}{{(e^x+e^y)}^2}\cdot \frac{e^y\cdot e^x}{{(e^x+e^y)}^2}=
=\frac{e^{2x}\cdot e^{2y}}{{(e^x+e^y)}^4}=
={(\frac{e^{x}\cdot e^{y}}{{(e^x+e^y)}^2})}^2=
={(z''_{xy})}^2
We got to the right side as required.
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