# Find the second-order partial derivatives of the given function If u=\ln (x^2+y^2)^{\frac{1}{2}}...

## Question:

Find the second-order partial derivatives of the given function

If {eq}u=\ln (x^2+y^2)^{\frac{1}{2}} ,(x,y) \neq 0 {/eq} , show that {eq}u_{xx}+u_{yy}=0 {/eq}

## Partial Derivatives:

Let {eq}z=f(x,y) {/eq} be a function of two variables then the partial derivatives of f with respect to x by taking other variable constant is denoted and defined by

{eq}f_{x}=\frac{\partial f}{\partial x}=\lim_{h\rightarrow 0}\frac{f(x+h,y)-f(x,y)}{h} {/eq}. Similarly, the partial derivative of f with respect to y can be defined.

This function can be extended to more than 2 variables.

## Answer and Explanation:

Consider the given function {eq}u=\ln(x^2+y^2)^{\frac{1}{2}}=\frac{1}{2}\ln(x^2+y^2) .........(1) {/eq}

Differentiating (1) partially with respect to x we get

{eq}u_{x}=\frac{1}{2}\times \frac{1}{x^2+y^2}(2x)=\frac{x}{x^2+y^2} ......(2) {/eq}

Differentiating (2) with respect to x we get

{eq}u_{xx}=\frac{(x^2+y^2)(1)-x(2x)}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2} ......(3) {/eq}

Now, using the symmetry relation

{eq}u_{yy}=\frac{x^2-y^2}{(x^2+y^2)^2} ......(4) {/eq}

Adding equation (3) and (4) we get

{eq}u_{xx}+u_{yy}=\frac{1}{(x^2+y^2)^2}(y^2-x^2+x^2-y^2)=0 {/eq}

#### Learn more about this topic: Solving Partial Derivative Equations

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 1
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