fbpx
calculus online - Free exercises and solutions to help you succeed!

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.

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply