N-th Derivative of Special Function

MIT has a cute little problem in their calculus problem set, which you can see here. I like this problem because it takes something complicated and shows that it actually reduces down to something very simple.

To start this problem off let’s suppose we have a function that we can write as the product of two other functions, y = u(x)v(x) . There are many such functions. For example, y(x) = x^2sin(x) .

To find the derivative we would need to use the power rule. To get warmed up let’s take two derivatives of y = u(x)v(x) .

y' = u'v + uv' (that’s just the product rule definition).

Applying the rule again we get four terms, two from the first product and two from the second:

y'' = u''v +u'v'  + u'v' + uv''

We can combine terms to get:

y'' = u''v + 2u'v' + uv'' 

This process can continue and indeed the problem set gives us Libniz’s formula for the n-th derivative, which is:

y^{(n)} = u^{(n)}v + \binom{n}{1}u^{(n-1)}v^{(1)} + \binom{n}{2}u^{(n-2)}v^{(2)} + \cdots + uv^{(n)}

This formula may look daunting at first, but it’s really pretty simple. Note that we are not raising our functions to powers, u^{(n-1)} means the (n - 1) -st derivative. Let’s apply Libniz’s formula to the second derivative of y(x) above.

y^{(2)} = u^{(2)}v + \binom{2}{1}u^{(2-1)}v^{(1)} + uv^{(2)}

y^{(2)} = u^{(2)}v + \frac{2!}{1!(2-1)!}u^{(2-1)}v^{(1)} + uv^{(2)}

y^{(2)} = u^{(2)}v + 2u^{(1)}v^{(1)} + uv^{(2)}

This is exactly what we got above, the only difference is the notation.

The problem now asks us to calculate y^{(p+q)} if y = x^p(x+a)^q .

Before we begin let’s make a few observations that will help us along the way. First, in this example  our u(x) is x^p and v(x) is (x+a)^q .

Next, let me remind you about two well known facts of x^p. As an example, I’ll take p to be 3 . Then we just have the following, where I will adopt the same notation used in Libniz’s formula.

y = x^3

y ^{(1)}= 3 \cdot x^2

y ^{(2)}= 3 \cdot 2 \cdot x^1

y ^{(3)}= 3 \cdot 2 \cdot 1

y ^{(4)}=  0

So what have we learned? Well, first we know that all derivatives higher than order p are 0. In our example, p was 3, and by the fourth derivative we got 0. Of course the derivative of 0 is 0, so from here on out (for all derivatives of order greater than 3), we’re going to get 0.

The second thing to notice is that the p-th derivative is actually p! . Again, in this case p was 3 and for y ^{(3)} our answer was 3 \cdot 2 \cdot 1 , otherwise known as 3! .

These two rules also hold for (x+a)^q . True, we have to use the chain rule, but of course the derivative of x+a is just 1.

Given what we’ve learned let’s move forward writing out the first few terms of Leibniz’s formula with our given y(x) .

y^{(p+q)} =x^{p^{(p+q)}}(x+a)^q + \binom{p+q}{1}x^{p^{(p+q-1)}}(x+a)^{q^{(1)}} + \binom{p+q}{2}x^{p^{(p+q-2)}}(x+a)^{q^{(2)}} + \cdots

So far all of these terms are 0 since the derivatives of x^p are of higher order than p . We can ask when they will stop being 0. This occurs on the (q+1) -st term when the derivative is p + q - q, which is just p . So we know that’s the first non-zero term. Let’s start with that term and write out the next few terms.

y^{(p+q)} =\binom{p+q}{q}x^{p^{(p+q-q)}}(x+a)^{q^{(q)}} + \binom{p+q}{q+1}x^{p^{(p+q-(q+1))}}(x+a)^{q^{(q+1)}} +\binom{p+q}{q+2}x^{p^{(p+q-(q+2))}}(x+a)^{q^{(q+2)}} + \cdots

Notice what happens here. For derivatives of order higher than q the (x+a)^q term is 0. So we are left with only a single term! That term is:

y^{(p+q)} =\binom{p+q}{q}x^{p^{(p+q-q)}}(x+a)^{q^{(q)}}

But this can be simplified using our rules from above. We can replace x^{p^{(p+q-q)}} with p! since this is the p-th derivative of x^p and (x+a)^{q^{(q)}} with q! since it’s the q-th derivative of (x+a)^q .

y^{(p+q)} =\binom{p+q}{q}p!q!

And now we can expand the binomial term. Recall that \binom{n}{k} = \frac{n!}{k!(n-k)!} .

y^{(p+q)} = \frac{(p+q)!}{q!(p+q-q)!}p!q!

y^{(p+q)} = \frac{(p+q)!}{q!p!}p!q!

y^{(p+q)} =(p+q)!

See what I mean! The derivatives of the product of two fairly general functions just turns into a few multiplications.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

w

Connecting to %s