A coincidence?

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Check out this dialogue between two calculus students (based on a true story):

Devyn: Yo Riley, I was looking at how the product rule works with more than two functions
Riley: What do you mean?
Devyn: Well, look at this: $\dd {x} f(x)g(x) h(x) = f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x)$
Riley: So?
Devyn: But, if $f(x)=g(x)=h(x)$ , then each of the terms looks the same: $\dd {x} f(x) f(x) f(x) = 3 f'(x) f(x) f(x)$
Riley: Where are you going with this?
Devyn: Well, I wonder what would happen if you tried to find the derivative of the product of many copies of the same function, would it be the same?
Riley: I’m not sure?
Devyn: Did you notice this: $f(x) f(x) f(x) = [f(x)]^3$ and $\begin{align*} \ddx f(x) f(x) f(x) &= \ddx [f(x)]^3\\ &= 3 f'(x) f(x) f(x) \\ &= 3 [f(x)]^2 f'(x) \end{align*}$
Riley: So?
Devyn: Do you think there is some way to use the derivative of $x^3$ to do this problem?
Riley: Maybe? Maybe there is a formula for this, just like there seems to be a formula for everything?

Let’s look at the following function and try to compute its derivative:

$f(x)=(x^{3})^{5}$

Based on what we have seen so far, perhaps we should just perform some algebra first before finding the derivative:

$\begin{align*} f(x) &= (x^{3})^{5}\\ &=x^{15} \\ \end{align*}$

$\begin{align*} \ddx f(x) &= \ddx x^{15}\\ &=15x^{14} \\ \end{align*}$

Now, let’s make a different computation:

$\ddx x^{3} = 3x^{2}$ and $\ddx x^{5} = 5x^{4}$

$\begin{align*} 5(x^{3})^{4} \cdot 3x^{2} &= 5x^{12} \cdot 3x^{2}\\ &=15x^{14} \\ \end{align*}$

In other words, if $g(x)=x^{5}$ and $h(x)=x^{3}$ , then

$\ddx g(h(x))=g'(h(x)) \cdot h'(x)$

Is this just a coincidence? No, it is not a coincidence. It turns out that this is a consequence of another rule of differentiation: The Chain Rule.