Subject: Calculus

Leibnizs Rule

It has been said that the "reverse method" of differentiation is integration, hence the name anti-differentiation. But what if we have these two processes acting both on a function? Does this mean that the two cancels each other leaving only the function alone? Well, that's not the case but we can make something more interesting about that fact.

Leibniz, considered by many as the founder of Calculus, makes such fact interesting by drawing a direct relation between integration and anti-differentiation when these are simultaneously being applied on a function. And this relation, which is also known as known as differentiation under the integral sign, is called the Leibniz's Rule which are as follows.

Leibniz's Rule

Suppose that f is a well-behaved and differentiable function and that f and its derivative is continuous over the region of interest, then

\frac{d}{dx}\int_{a}^{b} f(x,y)dy = \int_{a}^{b}\frac{\partial }{\partial x}f(x,y)dy

So what really the rule means is that no matter which comes first, either differentiation or integration, the result is still the same. In a sense, Leibniz's Rule just expresses the fact that integration and differentiation are, in a sense, commutative.

I will not buy into it. Can you give me an illustration?

Okay, to give a clearer picture into this stuff, let's crank up our head a bit and do this simple exercise. Let's choose one function, say f(x,y) = x^2 +2xy +y^2 and we'll do differentiation and then integration. On the other hand, let's do the same but this time, let's start first with integration then differentiation afterwards. We'll randomly choose a = 0 and b =5 as our integration limits.and let's see if both methods yields the same result as what the Leibniz Rule is trying to say. Well, they should be equal cause the rule says so, but it would be best if will check it ourselves.


Method A (differentiate and then integrate):

f(x,y) = x^2 +2xy +y^2
\frac{\partial}{\partial x}f(x,y) = D_{x}(x^2 +2xy +y^2)
\frac{\partial}{\partial x}f(x,y) = 2x +2y

Now, we have finished differentiating, so let's integrate...

\int_{0}^{5} \frac{\partial}{\partial x}f(x,y)dy = \int_{0}^{5} (2x +2y)dy
= 2xy + y^2|_{0}^{5} = 10x + 25

Then, we'll do the other method..

Method B (integrate and then differentiate):

f(x,y) = x^2 +2xy +y^2
\int_{0}^{5} f(x,y)dy = \int_{0}^{5} (x^2 +2xy +y^2)dy
=( x^{2}y +xy^{2} +\frac{1}{3}y^3)|_{0}^{5} = 5x^2 + 25x + \frac{1}{3}5^3
\frac{d}{dx}\int_{0}^{5} f(x,y)dy=D_{x}(5x^2 + 25x + \frac{1}{3}5^3)
=10x + 25 + 0

This is exactly identical to the results we got it method B of course. Thus, the Lebniz's Rule holds very much true in this case and should hold true for all other cases.

Other Important Case:

In our discussion above, we assume that the limits of integration, a and b, are constants. However, if a and b are functions of the differential variable x such that a(x) and b(x) are the limits of integration, then the Leibniz's Rule can be rewritten into a new form;

\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,y)dy = \int_{a(x)}^{b(x)}\frac{\partial }{\partial x}fdy + f(b(x), x)\frac{\partial b}{\partial x} - f(a(x), x)\frac{\partial a}{\partial x}

Example #1

Suppose you are to evaluate \frac{d}{dx}\int_{2x}^{3} f(x,y)dy , what form of Leibniz's Rule are you going to follow?

ANSWER: \frac{d}{dx}\int_{a}^{b} f(x,y)dy = \int_{a}^{b}\frac{\partial }{\partial x}f(x,y)dy

Example #2

If you are to calculate directly \frac{d}{dy}\int_{g(y)}^{h(y)} f(x,y)dx , what form of Leibniz's Rule is the most practical?

ANSWER: \frac{d}{dx}\int_{a(x)}^{b(x)} f(x,y)dy = \int_{a(x)}^{b(x)}\frac{\partial }{\partial x}fdy + f(b(x), x)\frac{\partial b}{\partial x} - f(a(x), x)\frac{\partial a}{\partial x}

Note: the variables x and y are interchanged in this case.

Example #3

\frac{d}{dy}\int_{1}^{2} xy^2 dx =3y, show that by using the alternate Leibniz form \int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx will still yield the same result.

\int_{1}^{2}\frac{\partial }{\partial y}xy^2dy

= \int_{1}^{2}2yxdy





Example #4

What happens to the Leibniz form \int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx when a=b?

\int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx = 0

Example #5

What happens if the function f under the integral sign is independent of either x or y?

\int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx = \int_{a}^{b}\frac{\partial }{\partial x}f(x,y)dy = 0

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