Welcome back to Calculus! In Section 3-4, we arrive at one of the most critical techniques in your mathematical toolkit: The Chain Rule. Up until now, we've dealt with relatively simple functions. But what happens when functions are nested inside one another? That is where the Chain Rule shines.

1. Understanding Composite Functions

The Chain Rule is used for composite functions, denoted as $F(x) = f(g(x))$. Think of these like layers of an onion or Russian nesting dolls. You have an "outside" function, $f$, and an "inside" function, $g$.

The formal definition of the derivative is:

$$ F'(x) = f'(g(x)) \cdot g'(x) $$

In plain English: Take the derivative of the outside function (keeping the inside exactly the same), and multiply it by the derivative of the inside function.

2. The General Power Rule

The most common application of the Chain Rule is when a polynomial is raised to a power. In the class notes, we looked at this example:

$$ y = (x^3 - 1)^{100} $$

Here, the "outside" function is the power of 100, and the "inside" function is $(x^3 - 1)$. Applying the rule:

  1. Differentiate the outside: Bring down the 100 and subtract 1 from the exponent. Keep $(x^3 - 1)$ untouched.
  2. Multiply by the derivative of the inside: The derivative of $x^3 - 1$ is $3x^2$.
$$ y' = 100(x^3 - 1)^{99} \cdot (3x^2) $$ $$ y' = 300x^2(x^3 - 1)^{99} $$

3. Handling Radicals

Don't let square roots or cube roots scare you. The key is to rewrite them as fractional exponents before you differentiate. As seen in the notes:

  • If $r(x) = \sqrt[3]{(3x^2+4x-2)^5}$, rewrite it immediately as: $$r(x) = (3x^2+4x-2)^{5/3}$$
  • If $f(x) = \frac{1}{\sqrt[3]{x^2+x+1}}$, rewrite using a negative exponent: $$f(x) = (x^2+x+1)^{-1/3}$$

Once rewritten, simply apply the Power Rule combined with the Chain Rule as usual.

4. Derivatives of General Exponentials ($b^x$)

We know that the derivative of $e^x$ is just $e^x$. But what about $2^x$ or $5^x$? The notes derive this formula using natural logs. The rule for any base $b > 0$ is:

$$ \frac{d}{dx}(b^x) = (\ln b) \cdot b^x $$

If we combine this with the Chain Rule (where the exponent is a function, like $5^{x^2}$), the formula becomes:

$$ \frac{d}{dx}(b^{g(x)}) = (\ln b) \cdot b^{g(x)} \cdot g'(x) $$

5. Combining Rules

Calculus gets messy when we mix rules. You will encounter problems requiring the Product Rule or Quotient Rule alongside the Chain Rule. Organization is key here!

For example, when differentiating $y = (2x+1)^5(x^3-x+1)^4$, you must use the Product Rule first ($f'g + g'f$), but calculating $f'$ and $g'$ will require the Chain Rule individually. Take it step-by-step and use plenty of paper!

6. The "Nested" Chain Rule

Finally, we looked at functions nested three layers deep, such as $f(x) = \sin(\cos(\tan x))$. You simply apply the chain rule repeatedly, working from the outside in:

$$ f'(x) = \cos(\cos(\tan x)) \cdot [-\sin(\tan x)] \cdot \sec^2(x) $$

Keep practicing these concepts. The Chain Rule is fundamental to everything we will do next in Calculus!