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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