Calculating Differential – Exercise 4233

Exercise

Find the differential of the function

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

Final Answer

Show final answer

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

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

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

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