# Calculating Differential – Exercise 4233

Exercise

Find the differential of the function

$$p=\ln \sqrt{x^2+y^2}$$

$$dp=\frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy$$

Solution

We will find the function differential with the differential formula

$$dp=p'_x dx+p'_y dy$$

In the formula above we see the function partial derivatives. Hence, we calculate them.

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

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

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

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

Now, we put the derivatives in the formula and get

$$dp=p'_x dx+p'_y dy$$

$$dp=\frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy$$

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