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First-Order Partial Derivative Of Functions With Three Variables

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EXAMPLE 1

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

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

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