Exercise
Find the differential of the function
p=\ln \sqrt{x^2+y^2}
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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