The Product Rule Really is Intuitive

One of the great geniuses of mathematics is Gottfried Liebniz. He discovered calculus along with Newton, and created the superior form of notation that we use today. He also flushed out the basics of calculus. And when he wasn’t doing calculus, he was working on philosophy. Liebniz was one of the greatest philosophers of the Enlightenment. But even the greatest geniuses stumble. Einstien divided by zero in his theory of relativity and missed out on predicting the Big Bang. And Liebniz got stuck for years because he expected that the derivative of u(x)v(x) should be u’(x)v’(x). Unfortunately the real answer is a lot more complicated:

derivative of u(x)v(x) = u’(x)v(x) + u(x)v’(x)

Apparently Liebniz got the correct answer but didn’t believe it was right. So he spent years trying to redo the problem before he accepted the truth. However, here we will learn why the real answer makes all the intuitive sense in the world.

When you multiply two lengths together you get a rectangle. So don’t think about u(x)v(x) as one value, think of it as a rectangle. Now consider what happens to the change of the area of the rectangle as u(x) and v(x) change, which is shown in the picture below.

The change in area = u’(x)v(x) + u(x)v’(x) + u’(x)v’(x)

However, when the change of x is small, which is the case when you are taking a derivative, then the far diagonal corner of the rectangle effectively disappears. So the last term of u’(x)v’(x) goes to zero. So you are left with the product rule:

derivative of u(x)v(x) = u’(x)v(x) + u(x)v’(x)

A Simple Test

Lets try a simple test of the product rule. Find the derivative of (2x + 1)(3x + 4) using the product rule, then by expanding and taking the power rule of the terms.

Product rule = 2(3x + 4) + (2x + 1)3 = 12x + 11

Now using the power rule. First we expand:

(2x + 1)(3x + 4) = 6x² + 11x + 4

Then apply the power rule to get:

12x + 11

Both answers are the same. So we can use the the product rule.

The Quotient Rule

The Product Rule also leads directly to the quotient rule.

v(x)÷u(x) = v(x)u(x)^-1

Then apply the product rule and the chain rule:

v’(x)u(x)^-1 + v(x)u’(x)(-1)u(x)^-2

     v’(x)u(x) – v(x)u’(x)
= ——————————-
        u(x)²

Fractional Exponents

If you are comfortable with implicit differentiation then you can now handle fractional exponents.

y = xⁿ = x^p/q
y^p = x^q

This is the essential insight – rewrite xⁿ as x^p/q and then take the qth power of each side. At that point it is just a matter of taking the derivative of both sides (using implicit differentiation) and then collecting the terms. Ultimately you get the expected answer, which that the derivative of x^p/q is ^p⁄_qx^{p/q – 1}