Calculus (X): How To Really Understand The Quotient Rule

Welcome to the tenth entry in the calculus series, where we cover the first principles behind the quotient rule of differentiation.

In my previous essay, I covered the product rule. Now, we are about to continue from where we ended the last time around.

Suppose that we are dealing with the following functions:

u(x) = 6x² + a

v(x) = 2x + b

y(x) = (6x² + a)/(2x + b) = u/v

dy/dx = ?? [a and b are constants]

Similar to the last time, u(x) and v(x) are separate functions of the independent variable ‘x’. We get y(x) by dividing u(x) by v(x). The question now is how do we find the derivative dy/dx? Let us find out.

Can We Obtain dy/dx Intuitively?

In my previous essay, we saw that an intuitive approach worked when y(x) was simply the product of u(x) and v(x). First, we differentiated u(x) separately with respect to ‘x’. Then, we differentiated v(x) with respect to ‘x’. Finally, we just multiplied these two results to get dy/dx.

In the case of division, this approach is not going to help us, since it is not clear how we can divide the numerator by the denominator. In other words, an intuitive method for our problem does not exist.

So, let us approach the problem at hand using first principles.

This essay is supported by Generatebg

---

Applying First Principles to Obtain the Quotient Rule

As we have been doing previously, let us increase ‘x’ by an infinitesimal amount of ‘dx’. Consequently, ‘y’ increases by ‘dy’, ‘u’ increases by ‘du’, and ‘v’ increases by ‘dv’ (where ‘dy’, ‘du’, and ‘dv’ are infinitesimal amounts). The resulting expression is as follows:

Let us start solving the right-hand side of the above equation by applying algebraic division (long division for polynomials) as follows:

We can use the same algorithm to continue solving the consequent remainder expressions:

The term in the last remainder you see is of the second degree smallness. So, we can neglect it and consider the quotient as our solution. Let us see what happens as we continue simplifying the expression from the quotient:

There we go. The final expression we get gives us the quotient rule of differentiation.

---

Interpreting the Quotient Rule of Calculus

Here is a step-wise summary of how you can interpret/execute the quotient rule:

1. Multiply the denominator function with the derivative of the numerator function.

2. Multiply the numerator function with the derivative of the denominator function.

3. Subtract the result from step no. 2 from the result from step no. 1.

4. Divide this result by the square of the denominator function to get the derivative.

Let us now apply what we have seen/learnt so far to solve our original example problem:

There we go; we have successfully applied the first principles to understand and execute the quotient rule of differentiation.

Up next in the series, I am considering working out some example problems to drive home the topics that I have covered so far.

---

Reference and credit: Silvanus Thompson.

If you’d like to get notified when interesting content gets published here, consider subscribing.

Further reading that might interest you:

• How To Casually Guess Numbers After Dice Throws?
• The New Industrial Revolution Is Here.
• The Hindsight Bias: Cause And Effect.

If you would like to support me as an author, consider contributing on Patreon.