# 10. Partial Derivatives

by M. Bourne

So far in this chapter we have dealt with **functions of single variables** only. However, many functions in mathematics involve 2 or more variables. In this section we see how to find derivatives of functions of more than 1 variable.

This section is related to, but is not the same as Implicit Differentiation that we met earlier.

### Example 1 - Function of 2 variables

Here is a function of 2 variables, *x* and *y*:

F(x,y) =y+ 6 sinx+ 5y^{2}

To plot such a function we need to use a 3-dimensional co-ordinate system.

## Partial Differentiation with respect to *x*

"Partial derivative with respect to *x*" means "regard all other letters as constants, and just differentiate the *x* parts".

In our example (and likewise for every 2-variable function), this means that (in effect) we should turn around our graph and look at it from the far end of the *y*-axis. We are looking at the *x*-*z* plane only.

We see a sine curve along the *x*-axis and this comes from the "6 sin *x*" part of our function *F*(*x*,*y*) = *y* + 6 sin *x* + 5*y*^{2}. The *y* parts are regarded as constants (in fact, 0 in this case).

Now for the **partial derivative** of

F(x,y) =y+ 6 sinx+ 5y^{2}

with respect to *x*:

`(del F)/(del x)=6 cos x`

The derivative of the 6 sin *x* part is 6 cos *x*. The derivative of the *y*-parts is zero since they are regarded as constants.

Notice that we use the curly symbol ∂ to denote "partial differentiation", rather than "`d`" which we use for normal differentiation.

**NOTE:** You can explore this example using this 3D interactive applet in the Vectors chapter. In the drop-down list of examples, this is the last one.

## Partial Differentiation with respect to y

The expression

Partial derivative with respect to

y

means

"Regard all other letters as constants, just differentiate the

yparts".

As we did above, we turn around our graph and look at it from the far end of the *x*-axis. So we see (and consider things from) the *y*-*z* plane only. (Notice the horizontal axis on the following graph is labeled *y*, not *x*.)

We see a parabola. This comes from the *y*^{2} and *y* terms in* F*(*x*,*y*) = *y* + 6 sin *x* + 5*y*^{2}. The "6 sin *x*" part is now regarded as a constant. (It actually has the effect of moving the base of the parabola `z=y+5y^2` down by 6 units.)

Now for the partial derivative of

F(x,y) =y+ 6 sinx+ 5y^{2}

with respect to *y*.

`(delF)/(dely)=1+10y`

The derivative of the *y*-parts with respect to *y* is 1 + 10*y*. The derivative of the 6 sin *x* part is zero since it is regarded as a constant when we are differentiating with respect to *y*.

**NOTE:** Once again, you can explore this particular example (rotate it, view it from different axes, etc) using the 3D interactive applet in the Vectors chapter. In the drop-down list of examples, select the last one.

## Second Order Partial Derivatives

We can find 4 different second-order partial derviatives. Let's see how this works with an example.

### Example 2

For the function we used above, *F*(*x*,*y*) = *y* + 6 sin *x* + 5*y*^{2}, find each of the following:

(a) `(del^2F)/(delydelx)`

Answer

This could also be written as

`del/(dely)[(delF)/(delx)]`

This expression means:

"First, find the partial derivative with respect to

xof the functionF(this is in brackets), then find the partial derivative with respect toyof the result ".

In our example above, we found

`(delF)/(delx)=6 cos x`

To find `(del^2F)/(delydelx)`, we need to find the partial derivative with respect to *y* of `(delF)/(delx)`.

` (del^2F)/(delydelx)=del/(dely)[(delF)/(delx)] `

`=del/(dely)[6 cos x]`

`=0`

Since cos *x* is a constant (when we are considering differentiation with respect to y), its derivative is just 0.

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(b) `(del^2F)/(delxdely)`

Answer

This could also be written as

`del/(delx)[(delF)/(dely)]`

This expression means

Find the partial derivative with respect to

xof the partial derivative with respect toy.

In our example above, *F*(*x,y*) = *y* + 6 sin *x* + 5*y*^{2}, we found

`(delF)/(dely)=1+10y`

To find `(del^2F)/(delxdely)`, we need to find the partial derivative with respect to *x* of `(delF)/(dely)`.

`(del^2F)/(delxdely)=del/(delx)[(delF)/(dely)]`

`=del/(delx)[1+10y]`

`=0`

Since *y* is a constant (when we are considering differentiation with respect to *x*), its derivative is just 0.

Easy to understand math videos:

MathTutorDVD.com

(c) `(del^2F)/(delx^2)`

Answer

This could also be written as

`del/(delx)[[delF)/(delx)]`

This expression means

Find the partial derivative with respect to

xof the partial derivative with respect tox.

In our example above, we found

`(delF)/(delx)=6 cos x`

To find `(del^2F)/(delx^2)`, we need to find the derivative with respect to *x* of `(delF)/(delx)`.

`(del^2F)/(delx^2)=del/(delx)[[delF)/(delx)]`

`=del/(delx)[6 cos x]`

`=-6 sin x`

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(d) `(del^2F)/(dely^2)`

Answer

This could also be written as

`del/(dely)[(delF)/(dely)]`

This expression means:

Find the partial derivative with respect to

yof the partial derivative with respect toy.

In our example above, *F*(*x*,*y*) = *y* + 6 sin *x* + 5*y*^{2}, we found

`(delF)/(dely)=1+10y`

To find `(del^2F)/(dely^2)`, we need to find the derivative with respect to *y* of `(delF)/(dely)`.

` (del^2F)/(dely^2)=del/(dely)[(delF)/(dely)]`

`=del/(dely)[1+10y]`

`=10`

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