Calculating Differential – Exercise 4239

Exercise

Find the differential of the function

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

Final Answer

Show final answer

Solution

We will find the function differential with the differential formula

$du=u'_x dx+u'_y dy+u'_z dz$

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

$u'_x(x,y,z)=-\frac{z}{{(x^2+y^2)}^2}\cdot 2x=$

$=\frac{-2xz}{{(x^2+y^2)}^2}$

$u'_y(x,y,z)=-\frac{z}{{(x^2+y^2)}^2}\cdot 2y=$

$=\frac{-2yz}{{(x^2+y^2)}^2}$

$u'_z(x,y,z)=\frac{1}{x^2+y^2}$

Now, we put the derivatives in the formula and get

$du=u'_x dx+u'_y dy+u'_z dz$

$du=\frac{-2xz}{{(x^2+y^2)}^2}dx+\frac{-2yz}{{(x^2+y^2)}^2}dy+\frac{1}{x^2+y^2}dz$

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