Exercise
Find the differential of the function
u=\frac{z}{x^2+y^2}
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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