Revisiting partial derivatives (Using Stochastic article 18 of n)

January 13th, 2010

(Read from the beginning) OR
(If you have not read article "17 of n" already then please consider reading it before continuing here

Reinforcing the concept of partial derivatives

It seems there are two things in calculus that are vital in understanding the derivation of ITO’s lemma

One of them is Taylor series, which I have covered in my previous post.

The other one is partial derivatives. Though I had covered this topic in one of my previous posts, I was not very happy with what I had done. So I decided to spend more time to make sure that I really understand this concept.

Let us take one more theoretical example:

Find the partial derivatives for f(x,y) = $xy^3 + y^2 + y$

[f(x,y) is just another way of saying function of x and y]

First let us keep x constant and find partial derivative with respect to y

Step 1: Say we need to find derivative of $3y^3$ which is $9y^2$ . Now instead of $3y^3$ , we need to find derivative of $xy^3$ . We treat x as constant (same as 3 in $3y^3$ ) and hence the partial derivative is $3xy^2$

Step 2: We already know that derivative of $y^2$ is 2y

Step 3: We also know that the derivative of y is 1.

Hence

$\frac{{\partial f}}{{\partial y}} = 3xy^2 + 2y + 1$

Now let us keep y constant and find partial derivative with respect to x.

Step 1: partial derivative of $xy^3$ is $y^3$ (derivative of x is 1 and $y^3$ is a constant)

Step 2: partial derivative of $y^2$ is 0 (Think over this, $y^2$ is a constant and hence derivative of $y^2$ with respect to x is zero)

Step 3: Same with y, the partial derivative with respect to x is zero.

Hence

$\frac{{\partial f}}{{\partial x}} = y^3$

You might be initially confused about why derivative of $y^2$ is zero. Look at this way. If we had a function $x^2 + 2x + 3$ then the derivative of this function will be 2x + 2. Notice that we have calculated derivative of 3 as zero since 3 is a constant. In our example $y^2$ is also a constant and hence the derivative of $y^2$ is zero.

The trick is to be able to mentally keep one or more of the variables as constant.

Let us now take a practical example to make sure we have really understood the concept:

Consider a rectangular box with width same as height. Let us say we denote width of the box as x and the length as y. Thus we get volume of the box as equal to $x^2 y$ (width * height * length).

Now let us say we keep length as constant.

(a) If the length is constant at 2 then in this case our formulae for volume is $2x^2$ . What is the rate of increase? Rate of increase is same as the first derivative i.e. 4x.

(b) Now let us keep length constant at 3. In this case our formulae for volume is $3x^2$ . What is the rate of increase? Rate of increase is 6x

(c) Finally let us keep length constant at 4. In this case our formulae for volume is $4x^2$ . What is the rate of increase? Rate of increase is 8x.

Is there any way to consolidate the above information? Yes, what we are basically doing above is to take partial derivatives i.e. keep one variable (length) constant.

Our original formulae for volume is $x^2y$ . i.e. $v = x^2 y$ . Now treat y as constant and take first derivative to find the rate of increase as 2xy. If you glance at the above calculations (a),(b),(c), you will realize that 2xy is just a fancy way of saying the same thing..4x, 6x, 8x etc (just substitute y with the actual values).

So now let us rewrite our fancy equation in the way mathematicians like it. It is very important to get used to this weird symbol.

$\frac{{\partial v}}{{\partial x}} = x^2y$

(I must say there seems to be something confusing over here. I would have thought that the partial derviative would have been shown like $\frac{{\partial x}}{{\partial v}}$ but I suppose there is some logical reason for showing it the other way i.e. $\frac{{\partial v}}{{\partial x}}$ )

(Continue reading: article 19 of n)

Wednesday, January 13th, 2010 | Stochastic calculus | Comments RSS | Respond | Trackback

2 Responses to Revisiting partial derivatives (Using Stochastic article 18 of n)

segaa says:

January 14, 2010 at 8:18 pm

>First let us keep y constant…
This eventually a misprint, we keep x constant here, not y.
I like reading your posts, thanks for all your great job!
harsh says:

January 15, 2010 at 7:02 pm

yes, you are right, thanx for pointing out the mistake. Corrected.

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Revisiting partial derivatives (Using Stochastic article 18 of n)

2 Responses to Revisiting partial derivatives (Using Stochastic article 18 of n)

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