# 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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