Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6467

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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