Calculus (V): How To Differentiate, Actually?

In calculus, learning how to differentiate is one of the most essential and fundamental tools of the trade. Differentiation is the fancy name given to the act of finding the value of the ratio dy/dx, where ‘y’ is the dependent variable and ‘x’ is the independent variable.

More specifically, we seek the value of the ratio of dy/dx in the limiting case where both ‘dy’ and ‘dx’ diminish to be infinitesimally small.

If you find all these words confusing, worry not; I will present a simple illustrated example in a forthcoming section.

In this essay, I will start with the conceptual explanation first and then present the illustrated example. We will be building slowly and steadily on what we already know.

This is the fifth essay in my calculus series. So, if you haven’t read my previous essays in this series, I’d recommend that you read them first. Let us begin by pondering why we bother computing dy/dx in the first place.

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Why Do We Need to Compute dy/dx?

Most calculus students take the ratio dy/dx for granted. The curious student might ask why we should bother. The answer lies at the heart of calculus. You see, calculus is the branch of mathematics that deals with change.

Some might define the ratio dy/dx as the slope of the change we are seeking to study. But I don’t wish to go that far.

The ratio dy/dx tells us what happens to ‘y’ when ‘x’ changes. That is, what happens to the dependent variable when the independent variable varies.

In my last essay on the notion of a derivative, I covered how ‘dy’ or ‘dx’ refers to an infinitesimally small fraction of ‘y’ or ‘x’. On the same vein, dy/dx tells us how ‘y’ varies with respect to ‘x’ for an infinitesimal variation in ‘x’.

In olden times, the ratio dy/dx was known as the differential coefficient. Thankfully, that name has died over the years. Today, we refer to this ratio as the derivative. Now that we have covered the derivative, why don’t we proceed to differentiation.

How to Differentiate? — The Concept

Consider the following function:

y = f(x) = x²

If we assume positive values for both ‘x’ and ‘y’, we see clearly that as ‘x’ increases, x² also increases. Consequently, ‘y’ also increases.

To realise this notion conceptually, let us increase ‘x’ by the infinitesimal fraction ‘dx’. As a result, ‘y’ also increases by the corresponding infinitesimal fraction ‘dy’ as follows:

y = x²

→ y + dy = (x + dx)²

When we apply the binomial theorem to the right-hand side, we get the following result:

y + dy = x² + 2xdx + (dx)²

Notice that the final term (dx)² is an infinitesimal of the second order of smallness.

The idea behind choosing ‘dx’ is such that any higher orders of smallness of the same fraction (such as (dx)², (dx)³, etc.) may be neglected. So, after neglecting this term, we get the following:

y + dy = x² + 2xdx

Now, let us subtract ‘y’ from both sides. Considering that y = x², we get:

y + dy – y = x² + 2xdx – x²

→ dy = 2xdx

Dividing both sides by ‘dx’, we get:

dy/dx = 2x

There we go. The ‘derivative’ of y = x² is 2x with respect to ‘x’. What does this mean? Well, why don’t we find out using a numeric example?

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How to Differentiate? — An Illustrated Example

Let us say that x = 1000. Consequently, y = x² = 1,000,000. Now, let us consider a negligible amount of ‘x’, say 1, and call this ‘dx’. That is, dx = 1. Then, as per our original equation, we compute the increased ‘y’ as follows:

y + dy = (x + dx)² = (1000 + 1)² = 1000² + 2*1000*1 + 1² = 1,002,001

If we choose to ignore the negligible term of second order (that is, 1²), we arrive at a (y + dy) value of 1,002,000. In other words, dy = 2000. Note that we considered a dx of 1 originally. Therefore:

dy/dx = 2000/1 = 2000 = 2*1000 = 2*x

What we have done here is illustrate how we could use the derivative to compute how ‘y’ will vary as ‘x’ varies. The astute reader might ask:

“Hang on a minute! You just conveniently ignored the ‘1’! Isn’t this cheating?”

Well, I hear you. Why don’t we change ‘dx’ to 1/10 or 0.1? Then, our original equation transforms as follows:

y + dy = (x + dx)² = (1000 + 0.1)² = 1000² + 2*1000*0.1 + 0.1² = 1,000,200.01

Now, the higher order term that we neglect becomes 0.01. Consequently, we can compute the derivative as follows:

dy/dx = 200/0.1 = 2000 = 2*1000 = 2*x

Note that even after, reducing ‘dx’ to 0.1, we arrive at the same result. The master stroke of calculus is to keep going lower and lower with ‘dx’ such that it converges on zero.

If you’ve read my previous essays, as soon as you read the word ‘converge’, another word should pop up in your head: limit!

Differentiation Using Limit Theory

Mathematicians use the Greek word Δ in place of ‘d’ to refer to infinitesimally small fractions of variables.

In limit theory, the derivative is computed as the limit of the ratio dy/dx as Δx diminishes to zero. The corresponding equation is as follows:

For the function f(x) = 8x, the above equation gives us the following result:

Δy/Δx = (8*Δx)/Δx

As Δx diminishes to zero, the ratio Δy/Δx gives the value of 8 as the limit. In other words, the derivative of y = f(x) = 8x is 8 with respect to ‘x’.

I hope this essay managed to give you a clearer picture of the fundamentals of differentiation. In my next essay, I will build upon what I have covered here and present another useful calculus tool: the power rule.

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Reference and credit: Silvanus Thompson.

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Further reading that might interest you:

• Is ‘0.99999…’ Really Equal To ‘1’?
• How To Benefit From Computer Science In Real Life (I)
• The Hindsight Bias: Cause And Effect

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