Power Rant

Power Rule §

The power rule is applicable to functions $x^n$ where $n$ is a real number. It essentially just states that the derivative of $y=x^n$, $y^{\prime }$, is equivalent to $n{x}^{x-1}$. Multiply the base of the function by the original power, then subtract one from the power.

It’s actually pretty intuitive when you consider the amount of polynomial multiplication you would have to do what you’re presented with monstrously big powers. A lot of people have trouble understanding where exactly this rule comes from.

To demonstrate, we’ll use our good old fashioned limit definition again to find $f^{\prime }(x)$ if $f(x)=x^6+x^5-x^3+9$.

$$\begin{array}{rl}{f}^{\prime }(x)& =\frac{f(x+h)-f(x)}{h}\\ & =\frac{\left[(x^6+x^5-x^3+9)+h\right]-(x^6+x^5-x^3+9)}{h}\\ & =\text{Lol you thought}\end{array}$$

Honestly just looking at that gives me a headache, so there was no way in hell I was going to type it all out. No, there’s a way better way to explain the power rule. For that, we’ll have to return to a Precalc Flashback™: Pascal’s Triangle.

Waffling 2: Even More Insane and Irrelevant Edition §

This is also almost 100% something we don’t have to know but I am bored and am going to write it out anyway.

Pascal’s Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it. It starts with a 1 at a top and expands downwards.

$$\begin{array}{ccccccccccc}& & & & & 1& & & & & \\ & & & & 1& & 1& & & & \\ & & & 1& & 2& & 1& & & \\ & & 1& & 3& & 3& & 1& & \\ & 1& & 4& & 6& & 4& & 1& \\ 1& & 5& & 10& & 10& & 5& & 1\\ ⋮& & ⋮& & ⋮& & ⋮& & ⋮& & ⋮\end{array}$$

In Precalculus, we used Pascal’s Triangle for binomial expansion. That was fun and all, but in Calculus we’re putting on our big boy gloves and applying it without thinking via the power rule.

Instead of looking at that long drudgery of an $f(x)$ defined above, we’ll go with a nice $g(x)=x^4$ to demonstrate.

$$\begin{array}{rl}{g}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{g(x+h)-g(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{(x+h{)}^4-x^4}{h}\end{array}$$

We could just multiply this out by hand, but it’s easier to just… apply Pascal’s Triangle to it.

$$\begin{array}{ccccccccc}1& & 4& & 6& & 4& & 1\end{array}$$

$$\begin{array}{rl}(x+h{)}^4& =1x^4+4x^{3}h+6x^2{h}^2+4h^{3}x+1h^4\end{array}$$

So, let’s chuck that back in there:

$$\begin{array}{rl}{g}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{(x^4+4x^{3}h+6x^2{h}^2+4h^{3}x+h^4)-(x^4)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\cancel{x^4}+4x^{3}h+6x^2{h}^2+4h^{3}x+h^4\cancel{-x^4}}{h}\\ & =\underset{h\to 0}{\lim }\frac{4x^{3}h+6x^2{h}^2+4h^{3}x+h^4}{h}\\ & =\underset{h\to 0}{\lim }4x^3+6x^{2}h+4h^{2}x+h^3\\ & =4x^3\end{array}$$

So $g^{\prime }(x)=4x^3$, or $g^{\prime }(x)=4(x^{4-1})$… the power rule. You can kind of see that, well, almost all of that algebra we did was totally senseless. Let’s look at it a more holistic way— yes, and we will bring back the damn triangle.

We go to the index $4$ row of the triangle and locate $4$ in the second and second to last positions. Remember that the first row of the triangle is $11$, not $1$. That is the ”$0$” index of the triangle.

$$\begin{array}{ccccccccccc}& & & & & 1& & & & & \\ & & & & 1& & 1& & & & \\ & & & 1& & 2& & 1& & & \\ & & 1& & 3& & 3& & 1& & \\ & 1& & 4& & 6& & 4& & 1& \\ 1& & 5& & 10& & 10& & 5& & 1\\ ⋮& & ⋮& & ⋮& & ⋮& & ⋮& & ⋮\end{array}$$

This is true for every value on the triangle except for $1$, which forms our outer layer and tip and all of that. We can also travel diagonally from the second and second to last positions to find the index values of other rows.

The value of $4$ first appears diagonally under $3$. What does that even mean? Well, we know that $3=4-1$ since we are not stupid. We also know that this means that a binomial with an exponent of ${}^3$ expanded yields a sequence of coefficients that includes $1,3,3,1$.

But what actually happens when we find the derivative? Going back to how we solved $g^{\prime }(x)$, we can see that we subtract the expansion of $(x+h{)}^4$ from $x^4$— removing our first value in the expansion of $(x+h{)}^4$, which is $x^4$ itself.

$$\underset{h\to 0}{\lim }\frac{\cancel{x^4}+4x^{3}h+6x^2{h}^2+4h^{3}x+h^4\cancel{-x^4}}{h}$$

The $x^4$ is essentially deleted from existence. We are going back a row, to a function where the base is raised to ${}^3$ instead of ${}^4$. But we are preserving the value of the original index of the row— $4$.