fbpx
calculus online - Free exercises and solutions to help you succeed!

Calculating Differential – Exercise 4239

Exercise

Find the differential of the function

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

Final Answer


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

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

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply