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


Given the differentiable function


Prove the equation



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)=



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)=



We will calculate the second order derivatives.

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


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


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






We got zero as required.

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