1. Break the Chain: Define $u$ as the inner function and $y$ as the outer function.
2. Differentiate the Links: Calculate $ rac{dy}{du}$ and $ rac{du}{dx}$ separately.
3. Connect: Multiply them together: $ rac{dy}{dx} = rac{dy}{du} · rac{du}{dx}$.
4. The Return: Replace the $u$ in your final answer with the original $x$ expression.
Break each function into its two component links ($u$ and $y$).
$y = \sqrt{x^3 + 1}$
$y = \cos(4x - 2)$
$y = e^{(x^2 + 5)}$
Using the links from Part I, calculate the derivatives $ rac{dy}{du}$ and $ rac{du}{dx}$.
For $y = \sqrt{u}$ and $u = x^3 + 1$:
For $y = \cos(u)$ and $u = 4x - 2$:
If you multiply $ rac{dy}{du} · rac{du}{dx}$, why does it equal $ rac{dy}{dx}$? What happens to the $du$'s visually? Does this mean the chain is getting "stronger" or just "connected"?
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Find the final $ rac{dy}{dx}$ by multiplying and re-substituting.
The Final Flow: Find $ rac{dy}{dx}$ for $y = \sqrt{x^3 + 1}$.
The Cosine Pulse: Find $ rac{dy}{dx}$ for $y = \cos(4x - 2)$.
Find $ rac{dy}{dx}$ for $y = \sin(\sqrt{x^2 + 10})$.
Link 1: $v = x^2 + 10$
Link 2: $u = \sqrt{v}$
Link 3: $y = \sin(u)$
Objective: Explain Leibniz notation to a younger sibling.
The Activity:
1. Make a chain of 3 paper clips.
2. Call the first one "The Handle" ($x$).
3. Call the middle one "The Bridge" ($u$).
4. Call the last one "The Toy" ($y$).
The Question: "If I pull the handle 2 inches, but the bridge is twice as big as the handle... how far does the toy move?" (4 inches).
The Lesson: "Total move = Handle move $·$ Bridge move. Every link multiplies the speed!"
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