$(3x + 1)^5$ is a function INSIDE a function: the inner function is $u = 3x + 1$, the outer function is $u^5$. Call the whole thing $y$. The chain rule says $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$, read aloud: dee y by dee x equals dee y by dee u, times dee u by dee x. Here $\dfrac{dy}{du}$ is the OUTER slope, the derivative of $u^5$ with respect to $u$, which is $5u^4$, read at the inner function's VALUE; and $\dfrac{du}{dx}$ is the INNER slope, the derivative of $3x + 1$, which is 3. A change in $x$ is scaled by the inside, then scaled again by the outside, and scalings multiply; on paper the $du$'s cancel as if they were numbers, which is how to remember the form. In function notation, for $y = f(g(x))$, the same rule reads $(f(g(x)))' = f'(g(x)) \cdot g'(x)$, f prime of g of x, times g prime of x: the outer derivative evaluated at the inner function, times the inner derivative.

At $x = 1$: $u = 4$, $\dfrac{dy}{du} = 5 \cdot 4^4 = 1280$, $\dfrac{du}{dx} = 3$, so $\dfrac{dy}{dx} = 1280 \cdot 3 = 3840$. Written out, $15(3x + 1)^4$ at 1 is $15 \cdot 256 = 3840$, the same. The trap in the lesson's name is dropping the inner derivative: $5(3x + 1)^4$ is the outer slope alone, as if the bracket were a plain $x$, and it is wrong by a factor of 3. The other slip is evaluating the outer derivative at $x$ instead of at $u$: $5x^4 \cdot 3$ raises the wrong thing to the fourth power.