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First-order partial derivative of functions with three variables

(Last Updated On: June 15, 2019)

First-order partial derivative of functions with three variables. Dear friends, today’s topic is the first-order partial derivative of functions with three variables.



So this is more like a re-visit to the good old topic. Have a look!!

First-order partial derivative of functions with three variables

In my earlier post on the first-order partial derivative of functions with two variables, I have discussed in great detail how things work.

So here without much ado, I’ll start with some examples.

Examples of the First-order partial derivative of functions with three variables

Disclaimer: None of these examples is mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here are the examples.

Example 1

According to Stroud and Booth (2013)*  “If V = x^2 + y^2 + z^2, express in its simplest form

    \[x\frac{\partial V}{\partial x} + y\frac{\partial V}{\partial y} + z\frac{\partial V}{\partial z}.\]

”

Solution

Here the given function is V = x^2 + y^2 + z^2.

And, I have to find out the value of x\cfrac{\partial V}{\partial x} + y\cfrac{\partial V}{\partial y} + z\cfrac{\partial V}{\partial z}.

So I’ll start with \cfrac{\partial V}{\partial x}.

Step 1

First of all, I’ll differentiate V partially with respect to x to get

    \[\frac{\partial V}{\partial x} = \frac{\partial }{\partial x}\left(x^2 + y^2 + z^2\right).\]

Thus I get

(1)   \begin{equation*} \frac{\partial V}{\partial x} = 2x. \end{equation*}

Next,  I’ll differentiate V partially with respect to y to get

    \[\frac{\partial V}{\partial y} = \frac{\partial }{\partial y}\left(x^2 + y^2 + z^2\right).\]

Therefore I get

(2)   \begin{equation*} \frac{\partial V}{\partial y} = 2y. \end{equation*}

Finally, I’ll differentiate V partially with respect to z to get

    \[\frac{\partial V}{\partial z} = \frac{\partial }{\partial z}\left(x^2 + y^2 + z^2\right).\]

Thus I get

(3)   \begin{equation*} \frac{\partial V}{\partial z} = 2z. \end{equation*}

So, now I’ll find out the value of x\cfrac{\partial V}{\partial x} + y\cfrac{\partial V}{\partial y} + z\cfrac{\partial V}{\partial z}.

Step 2

For that, I’ll use the values of \cfrac{\partial V}{\partial x}, \cfrac{\partial V}{\partial y} and \cfrac{\partial V}{\partial z} from equations (1), (2) and (3) respectively.

Thus it will be

    \[x\cfrac{\partial V}{\partial x} + y\cfrac{\partial V}{\partial y} + z\cfrac{\partial V}{\partial z}= x. 2x + y. 2y + z. 2z.\]

Therefore I get

    \begin{eqnarray*} x\frac{\partial V}{\partial x} + y\frac{\partial V}{\partial y} + z\frac{\partial V}{\partial z} &=& 2x^2+2y^2+2z^2\\ &=& 2(x^2+y^2 + z^2)\\ &=& 2V. \end{eqnarray*}

Hence I can conclude that this is the answer to this example.

Now I’ll go to the next example.

Example 2

According to Stroud and Booth (2013)* “If u = \cfrac{x+y+z}{(x^2+y^2+z^2)^{\frac{1}{2}}}, show that x \cfrac{\partial u}{\partial x}+ y \cfrac{\partial u}{\partial y} + z \cfrac{\partial u}{\partial z} = 0.”

Solution

In this example, the given function is u = \cfrac{x+y+z}{(x^2+y^2+z^2)^{\frac{1}{2}}}

And, I have to prove that x\cfrac{\partial u}{\partial x} + y\cfrac{\partial u}{\partial y} + z\cfrac{\partial u}{\partial z} = 0.

So I’ll start with \cfrac{\partial u}{\partial x}.

Step 1

First of all, I’ll differentiate u partially with respect to x to get

    \[\frac{\partial u}{\partial x} = \frac{\partial }{\partial x}\left(\cfrac{x+y+z}{(x^2+y^2+z^2)^{\frac{1}{2}}}\right)\\= \frac{(x^2+y^2+z^2)^{\frac{1}{2}}.1-\frac{1}{2}(x^2+y^2+z^2)^{-\frac{1}{2}}(2x)(x+y+z)}{x^2+y^2+z^2}.\]

Therefore I get

    \begin{eqnarray*} \frac{\partial u}{\partial x} &=& \frac{(x^2+y^2+z^2)^{\frac{1}{2}}-x(x^2+y^2+z^2)^{-\frac{1}{2}}(x+y+z)}{x^2+y^2+z^2}\\ &=& \frac{(x^2+y^2+z^2)-x(x+y+z)}{(x^2+y^2+z^2)(x^2+y^2+z^2)^{\frac{1}{2}}}\\ &=& \frac{1}{(x^2+y^2+z^2)^{\frac{1}{2}}}-\frac{x}{(x^2+y^2+z^2)}.u. \end{eqnarray*}

Thus I can say x\cfrac{\partial u}{\partial x} will be

(4)   \begin{equation*} x\frac{\partial u}{\partial x} = \frac{x}{(x^2+y^2+z^2)^{\frac{1}{2}}}-\frac{x^2u}{(x^2+y^2+z^2)}. \end{equation*}

In the same way I can also get the value of \cfrac{\partial u}{\partial y}.

Thus \cfrac{\partial u}{\partial y} will be

    \[\frac{\partial u}{\partial y}= \frac{1}{(x^2+y^2+z^2)^{\frac{1}{2}}}-\frac{y}{(x^2+y^2+z^2)}.u.\]

Therefore I can say y\cfrac{\partial u}{\partial y} will be

(5)   \begin{equation*} y\frac{\partial u}{\partial y} = \frac{y}{(x^2+y^2+z^2)^{\frac{1}{2}}}-\frac{y^2u}{(x^2+y^2+z^2)}. \end{equation*}

Similarly I can also get the value of \cfrac{\partial u}{\partial z}.

Thus \cfrac{\partial u}{\partial z} will be

    \[\frac{\partial u}{\partial z}= \frac{1}{(x^2+y^2+z^2)^{\frac{1}{2}}}-\frac{z}{(x^2+y^2+z^2)}.u.\]

Therefore I can say z\cfrac{\partial u}{\partial z} will be

(6)   \begin{equation*} z\frac{\partial u}{\partial z} = \frac{z}{(x^2+y^2+z^2)^{\frac{1}{2}}}-\frac{z^2u}{(x^2+y^2+z^2)}. \end{equation*}

So, now I’ll find out the value of x\cfrac{\partial u}{\partial x} + y\cfrac{\partial u}{\partial y} + z\cfrac{\partial u}{\partial z}.

Step 2

For that, I’ll use the values of x\cfrac{\partial u}{\partial x}, y\cfrac{\partial u}{\partial y} and z\cfrac{\partial u}{\partial z} from equations (4), (5) and (6) respectively.

Thus it will be

    \begin{eqnarray*} x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} &=& \frac{1}{(x^2+y^2+z^2)^{\frac{1}{2}}}(x+y+z)-\frac{u}{(x^2+y^2+z^2)}(x^2+y^2+z^2)\\ &=& (u-u)~~~~~~\text{since}~(x^2+y^2+z^2)\neq0\\ &=& 0. \end{eqnarray*}

Hence I can conclude that I have proved x\cfrac{\partial u}{\partial x} + y\cfrac{\partial u}{\partial y} + z\cfrac{\partial u}{\partial z} = 0.

This is the answer to this example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

*Reference: K. A. Stroud and Dexter J. Booth (2013): Engineering mathematics, Industrial Press, Inc.; 7th Edition (March 8, 2013), Chapter: Partial differentiation 1, Test exercise 14, p. 749, Q. No. 2(a) (Example 1); Further problems 14, p. 750, Q. No. 4 (Example 2).

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