Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6472

Exercise

Given the differentiable function

$z(x,y)=\ln(e^x+e^y)$

Prove the equation

$z'_x+ z'_y=1$

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 put the partial derivatives in the left side of the equation we need to prove.

$z'_x+ z'_y=$

$=\frac{e^x}{e^x+e^y}+\frac{e^y}{e^x+e^y}=$

$=\frac{e^x+e^y}{e^x+e^y}=$

$=1$

We got one as required.

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