Video Lecture

In this video we aim to develop a rule to quickly compute the derivative of $f(x)=x^n$ for any real number $n$. In particular we want to develop a rule that lets us avoid needing to compute the limit of the difference quotient each time!

Text and Additional Details

Often the best place to start is to just see what happens when we just go ahead and try to compute what we want. To this end, let’s see what happens when we try to determine the limit of the difference quotient for $f(x)=x^n$.

We’ve already run into a bit of trouble! Normally we would try to expand the $(x + \Delta x)^n$ term and then combine like terms. Unfortunately, since we don’t know what “$n$” actually equals, it’s hard to know how to go about this... if we knew $n$ was 2, or 3, or 4, or any specific number, we could do it (although it could get annoying for larger $n$ values certainly!) but the entire point is that we don’t know what $n$ is specifically... so what now?

Technically, what mathematicians use here, is something called the Binomial Expansion Theorem. The specifics of this theorem however are a little outside the scope of this course, so instead we will build a somewhat intuitive understanding of what it does and how we can use it (rather than rigorously proving the full result).

Binomial Expansion Theorem.... ish.

Let’s start by thinking about what $(x + \Delta x)^n$ means - the product of $x + \Delta x$ with itself $n$ times. So we can write it as: $$(x + \Delta x)^n = \underbrace {(x + \Delta x)\cdot (x + \Delta x)\cdot (x + \Delta x)\cdots (x + \Delta x)}_\text {`n' times}\cdot$$ When trying to expand this out we can think of it as choosing which of the two terms, the $x$ or the $\Delta x$ we want to multiply together from each set of parentheses; and the full expansion is multiplying together every possible combination of these choices. For example, let’s see what happens when $n=3$:

Here we can see that, if we select one from $x$ or $\Delta x$ in each set of parentheses and multiply them together, and do every possible combination exactly once, we get the full expansion (which we then need to combine back together by combining like terms). This is a bit of a bulky way to expand things of the form $(a + b)^n$, but it works. And we can make a few key observations that will be key for using this to solve our original problem.

The first observation is that, there will be exactly one combination that is entirely the first term to the power of $n$. In our example of $(x + \Delta x)^3$, there was exactly one $x^3$. Likewise, for $(x + \Delta x)^n$, there will be exactly one $x^n$, namely the combination where the $x$ is chosen in each set of parentheses to be multiplied together.

The second observation is that there will be exactly $n$ combinations that have a single $\Delta x$ factor. In our example, multiplying $(x + \Delta x)^3 = (\xI + \delxI )\cdot (\xII + \delxII )\cdot (\xIII + \delxIII )$ has, as factors with only a single $\Delta x$ factor, the following: $(\xI \cdot \xII \cdot \delxIII )$, $(\xI \cdot \delxII \cdot \xIII )$, $(\delxI \cdot \xII \cdot \xIII )$. Thus there are $3$ terms with a single $\Delta x$ factor for $(x + \Delta x)^3$, and similarly there are $n$ terms with exactly one $\Delta x$ factor in $(x + \Delta x)^n$. Notice that, since each term has a total of $n$ factors in it (whichever terms you selected to multiply together when you were selecting a term from each set of parentheses) this means that the $n$ terms with a single $\Delta x$ factor must then have $x^{n-1}$ as the remaining factors, since they are formed by choosing $n-1$ “$x$’s” and 1 $\Delta x$ from the $n$ sets of parentheses.

The final observation is that, all other terms have at least two $\Delta x$ factors. As we can see in our $(x + \Delta x)^3$ example, we had the single $x^3$ term, the 3 $x^2\cdot \Delta x$ terms, and then the rest of the terms either had the form $(x \Delta x)^2$ or $(\Delta x)^3$; both of which have at least two $\Delta x$ factors.

So, we can conclude then, that if we expand out $(x + \Delta x)^n$ we will get a single $x^n$, $n$ terms of the form $x^{n-1}\Delta x$, and the rest of the factors (whatever they are) will have at least 2 $(\Delta x)$ factors. Since these extra terms all have at least 2 $\Delta x$ factors, we can factor a $(\Delta x)^2$ out from these terms without any trouble. So, that means we can write $(x + \Delta x)^n$ as follows: $$(x + \Delta x)^n = x^n + nx^{n-1}(\Delta x) + (\Delta x)^2(\text {Leftover terms})$$ Where the leftover terms are all of the form $x$ to some non-negative power times $\Delta x$ to some non-negative power.

So, what were we doing again?

Now, believe it or not, we can return to our original problem! Recall our previous calculation, and apply our new result for expanding $(x + \Delta x)^n$:

Believe it or not, we’re done! We notice that, since the powers of $x$ and $\Delta x$ are all non-negative in the “leftover terms” part, it is bounded (i.e. not going to infinity) and so the factor of $\Delta x$ outside going to zero crushes the entire term down to zero, leaving just $nx^{n-1}$ left over. We can now formally record the polynomial rule.

Thus, after all this effort, we can conclude that for a general function $f(x) = x^n$, the derivative is simply $f'(x) = nx^{n-1}$. It took a surprising amount of effort to get there, but the beautiful thing is that, since we have proved the general result, we no longer need to go through this effort to do this derivative. Consider, for example, the time and effort involved in expanding something like $(x + \Delta x)^{10}$ in the difference quotient to find the derivative of $f(x) = x^{10}$... it would certainly be doable, but a tedious nightmare. Now we simply use our rule to conclude that the derivative is $f'(x) = 10x^9$. As we said, a little (or sometimes a lot) more work up front, to have much less work later!

Some useful consequences of the polynomial rule theorem are recorded next, but notice that these are just specific applications of the above theorem. There is no need to memorize these, as you get these results simply by applying the polynomial rule. Nonetheless these situations are so common (and not always intuitive) that they bear mentioning.

We have seen how we can make an argument to determine a nice and elegant rule to find the derivative for any function of the form $f(x)=x^n$. Strictly speaking the argument we gave is one that only works for $n$ that are non-negative integers, but actually this rule holds for any (finite) real number $n$, including irrational and negative values.

This means we can compute the derivative of something like $f(x) = x^{\sqrt {2}}$, which would simply be $f'(x) = \sqrt {2}x^{\sqrt {2}-1}$. Trying to compute this derivative using the difference quotient would be much more difficult than even one would think, but we will take it as proved for now. For those that are interested in the deeper complications in proving this kind of case (for non-integer values of $n$), students would be well advised to look into a real analysis course, like Advanced Calculus or Introduction to Modern Analysis.