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Calculating Differential – Exercise 4233

Exercise

Find the differential of the function

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

Final Answer


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

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