Calculus (VII): How To Differentiate Constants? - An illustration showing that d(ax²)/dx = 2ax. Below this equation, the following question is asked: "But How??"

Welcome to the seventh entry in my calculus series — an account on how to differentiate constants. My aim with the calculus series is to present the subject practically and comprehensibly with a strong emphasis on the ground fundamentals.

In my previous essay, I covered how we arrive at the power rule of calculus. In this essay, I will be formally extending this concept to treat constants in calculus.

In case this essay is your first exposure to my calculus series, I suggest you read the previous essays first to maximise your value gain. Without any further ado, let us begin.

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Constants in Calculus

So far, we have been working with variables. More specifically, we have been seeking the ratio of the proportion by which ‘y’ changes when ‘x’ changes (or we change it). Since the value of ‘y’ is dependent on ‘x’ we call ‘y’ the dependent variable and ‘x’ the independent variable.

The ratio we seek, dy/dx, is known as the derivative. It gives the proportion of change in ‘y’ for an infinitesimal variation in ‘x’. And we call the process of obtaining this ratio “differentiation”. So far, so good.

Now, constants are numbers that do NOT vary when ‘x’ and/or ‘y’ vary. In other words, the value of a constant remains unchanged regardless of the circumstances (hence ‘constant’).

But when constants occur in function expressions whose derivative we seek, we still have to differentiate them. Let us find out how we could do this using a simple example.

How to Differentiate Constants — Added Constants Vanish

Consider the following simple function expression:

y = f(x) = x² + 5

When we increase ‘x’ infinitesimally by ‘dx’, ‘y’ correspondingly increases infinitesimally by ‘dy’. Consequently, the new expression is as follows:

y + dy = (x + dx)² + 5

Let us now follow the same approach we followed in the previous essays and expand the right-hand side using the binomial rule:

y + dy = x² + 2xdx + (dx)² + 5

We can neglect higher powers of ‘dx’ due to the higher degree of ‘smallness’. In this case, we can drop (dx)². I have already covered why this possible in my previous essays. So, I am skipping the explanation here. The updated equation is as follows:

y + dy = x² + 2xdx + 5

Now, let us subtract y = (x² + 5) from both sides of the above equation:

y + dy − y = x² + 2xdx + 5 − x² − 5

→ dy = 2xdx

When we divide both sides by ‘dx’, we get the derivative:

dy/dx = 2x

The fact that the constant has vanished shows that it plays no role in the variation of ‘y’ with respect to ‘x’. When we generalise this result to any added constant, say, ‘a’, we see that it would vanish.

In other words, for y = (x^n + a), dy/dx = n*x^(n-1) (the product rule). This result applies to negative constants as well. Now that we have seen how to differentiate additive constants, let us proceed to multiplied constants next.

How to Differentiate Constants — Multiplied Constants Remain

To explore how to treat multiplied constants, consider the following expression:

y = 5x²

Applying the same procedure as before, we obtain the following result:

Calculus (VII): How To Differentiate Constants? — y + dy = 5(x + dx)²; y + dy = 5(x² + 2xdx + (dx)²); Neglecting 5*(dx)², we get: y + dy = 5 (x² + 2xdx); Subtracting y = 5x² from both sides: y + dy — y = 5x² + 10xdx — 5x²; dy = 10xdx; Dividing both sides by dx, we get the derivative: dy/dx = 10x;
d(5x²)/dx — Math illustrated by the author

When we generalise this result to the expression y = ax^n, we would see that the derivative takes the form: dy/dx = a*[n*x^(n-1)]. In other words, when x^n is multiplied by a constant, the derivative is the constant times the result of the power rule.

The generalisation we have just derived is also applicable for division, since it can be transformed into multiplication. For example, dividing by 5 is equivalent to multiplying by (1/5).

Closing Thoughts

Calculus practitioners might feel that the results that I have presented in this essay are trivial. However, in my experience, it is the simplest mathematical results that are most easily taken for granted. This is especially true for beginners.

It helps a great deal if a calculus student or practitioner is in a position to explain the fundamental mathematics behind the most basic calculus tools instead of saying “It just works that way.”

With this new tool under our belt, we are ready to tackle functions that are simple algebraic expressions. In the forthcoming essays, I will be covering combinations of two or more functions such as sums, differences, etc.

I have also been thinking of adding exercise-like sections to these essays. In case you would find such an approach beneficial, do let me know in the comments section.

Reference and credit: Silvanus Thompson.

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