Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6465

Exercise

Given the differentiable function

$z(x,y)=\ln\frac{1}{\sqrt{x^2+y^2}}$

Prove the equation

$z''_{xx}+z''_{yy}=0$

Proof

We will use the chain rule to calculate the partial derivatives of z.

$z'_x=\frac{1}{\frac{1}{\sqrt{x^2+y^2}}}\cdot\frac{-1}{{(\sqrt{x^2+y^2})}^2}\cdot\frac{1}{2\sqrt{x^2+y^2}}\cdot (2x)=$

$=\sqrt{x^2+y^2}\cdot\frac{-x}{x^2+y^2}\cdot\frac{1}{\sqrt{x^2+y^2}}=$

$=\frac{-x}{x^2+y^2}$

$z'_y=\frac{1}{\frac{1}{\sqrt{x^2+y^2}}}\cdot\frac{-1}{{(\sqrt{x^2+y^2})}^2}\cdot\frac{1}{2\sqrt{x^2+y^2}}\cdot (2y)=$

$=\sqrt{x^2+y^2}\cdot\frac{-y}{x^2+y^2}\cdot\frac{1}{\sqrt{x^2+y^2}}=$

$=\frac{-y}{x^2+y^2}$

We will calculate the second order derivatives.

$z''_{xx}=\frac{-(x^2+y^2)+x\cdot 2x}{{(x^2+y^2)}^2}=$

$=\frac{x^2-y^2}{{(x^2+y^2)}^2}$

$z''_{yy}=\frac{-(x^2+y^2)+y\cdot 2y}{{(x^2+y^2)}^2}=$

$=\frac{y^2-x^2}{{(x^2+y^2)}^2}$

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

$z''_{xx}+z''_{yy}=$

$=\frac{x^2-y^2}{{(x^2+y^2)}^2}+\frac{y^2-x^2}{{(x^2+y^2)}^2}=$

$=\frac{x^2-y^2+y^2-x^2}{{(x^2+y^2)}^2}=$

$=\frac{0}{{(x^2+y^2)}^2}=$

$=0$

We got zero as required.

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