Calculus (VI): How To Really Learn The Power Rule?

When it comes to differentiating low-order functions in calculus, the power rule is arguably the most frequently used instrument. In my previous essay in the calculus series, I covered the fundamentals behind the concept of differentiation.

In this essay, I will be building upon what we have covered so far and differentiate higher powers of ‘x’. By doing this, I will be illustrating how the so-called power rule of calculus emerges naturally.

The power rule is nothing but a generalisation we obtain by applying the process of differentiation to various powers of the independent variable ‘x’. Once you understand how the power rule works, you have a powerful tool in your calculus toolkit.

Just as a reminder, this is the sixth essay in my calculus series. If you have not read my previous essays, I would recommend reading them before you read this one. Without any further ado, let us begin.

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Revisiting Fundamentals

Previously, I showed how we arrive at the derivative of the function y = x². Why don’t we start from there?

Let us increase ‘x’ by the infinitesimal fraction ‘dx’. Correspondingly, ‘y’ also increases by the infinitesimal fraction ‘dy’. Then, we arrive at the following expression:

y = x²

→ y + dy = (x + dx)²

After applying the binomial theorem to the right-hand side, we arrive at the following result:

y + dy = x² + 2xdx + (dx)²

We choose dx such that (dx)² can be negligible. We may, therefore, drop terms of second-degree smallness and above ((dx)², (dx)³, etc) to get:

y + dy = x² + 2xdx

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

y + dy — y = x² + 2xdx — x²

→ dy = 2xdx

Dividing both sides by ‘dx’ gives us the value of the derivative of y = x² with respect to ‘x’ in the form of the ratio dy/dx:

dy/dx = 2x

So far, so good. Now, let us take this logic a bit further to higher powers of ‘x’.

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Differentiating Higher Powers of x

Let us now consider the function y = x³. Applying the same logic as before, as we increase ‘x’ by ‘dx’, we increase ‘y’ by ‘dy’. Applying the binomial formula to the resulting expression, we get a similar result here as well:

y = x³

→ y + dy = x³ + 3x²*dx + 3x*(dx)² + (dx)³

Since we choose ‘dx’ to be negligible, we may disregard even smaller terms in the form of (dx)², (dx)³, etc. As a result, the expression becomes:

y + dy = x³ + 3x²*dx

Let us now subtract y (= x³) from both sides of this expression:

y + dy — y = x³ + 3x²*dx — x³

→ dy = 3x²*dx

Dividing both sides by ‘dx’, we get the value of the derivative of y = x³ with respect to ‘x’ in the form of the ratio dy/dx:

dy/dx = 3x²

If you are interested, you may try applying the same procedure to y = x⁴. You would get the following result:

dy/dx = 4x³, for y = x⁴

We can clearly see a pattern emerging. Why don’t we try and generalise this result?

The Power Rule Emerges

The binomial formula for any exponent ’n’ is as follows:

When we plug in ‘x’ and ‘dx’ in place of ‘a’ and ‘b’ respectively, we get the following result:

There we go. This formula is known in calculus for the power rule. But before we celebrate, we have just established that this pattern works for positive whole numbers. We still need to check the pattern for fractional and negative exponents.

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Does the Power Rule Hold for Fractional and Negative Exponents?

Consider the function y = x^(1/2). By the same treatment as before:

As we can see, the power rule works for the fractional exponent. Now, let us check negative exponents by considering y = x^(-2):

There we go. We have now checked that the power rule works for negative exponents as well. So, the derivative of y = x^n for any ’n’ with respect to ‘x’ is given by the power rule as follows:

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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’?
• Learning Better Using Tacit Knowledge
• Why Is 3 A Special Denominator In Division?

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