Chain Rule for MATH 140

Exam Relevance for MATH 140

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The chain rule is the most important MATH 140 technique. Appears in nearly every derivative problem—master it completely.

Lesson

What is the Chain Rule?

The chain rule is how you differentiate composite functions — functions inside other functions.

Example of a composite function: $f(x) = (3x + 1)^5$

Here, there's an "outer function" (raising to the 5th power) and an "inner function" ($3x + 1$).

The problem: You can't just use the power rule directly. $(3x+1)^5 \neq 5(3x+1)^4$ — that's missing something!


The Formula

If $y = f(g(x))$, then:

$$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$$

In words: "Derivative of the outer (with the inner unchanged) times derivative of the inner."

Alternative notation (Leibniz): If $y = f(u)$ and $u = g(x)$:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$


How to Identify When You Need the Chain Rule

Ask yourself: "Is there a function inside another function?"

Expression Outer Function Inner Function Need Chain Rule?
$(2x+1)^3$ $(\cdot)^3$ $2x+1$ ✅ Yes
$\sin(x^2)$ $\sin(\cdot)$ $x^2$ ✅ Yes
$e^{5x}$ $e^{(\cdot)}$ $5x$ ✅ Yes
$x^3$ ❌ No (just power rule)
$\sin(x)$ ❌ No (inner is just $x$)

Example 1: Power of a Linear Function

Problem: Find $\frac{d}{dx}[(3x+1)^5]$

Step 1: Identify outer and inner functions:

  • Outer: $f(u) = u^5$
  • Inner: $u = g(x) = 3x + 1$

Step 2: Find each derivative:

  • $f'(u) = 5u^4$
  • $g'(x) = 3$

Step 3: Apply chain rule — outer derivative (keep inner unchanged) times inner derivative: $$\frac{d}{dx}[(3x+1)^5] = 5(3x+1)^4 \cdot 3$$

$$\boxed{15(3x+1)^4}$$


Example 2: Power of a Quadratic

Problem: Find $\frac{d}{dx}[(x^2 - 4x)^3]$

Step 1: Identify the parts:

  • Outer: $u^3$
  • Inner: $u = x^2 - 4x$

Step 2: Find derivatives:

  • Outer derivative: $3u^2$
  • Inner derivative: $2x - 4$

Step 3: Apply chain rule: $$\frac{d}{dx}[(x^2-4x)^3] = 3(x^2-4x)^2 \cdot (2x-4)$$

$$\boxed{3(x^2-4x)^2(2x-4)}$$

Or factored: $6(x^2-4x)^2(x-2)$


Example 3: Trig with Chain Rule

Problem: Find $\frac{d}{dx}[\sin(x^2)]$

Step 1: Identify the parts:

  • Outer: $\sin(u)$
  • Inner: $u = x^2$

Step 2: Find derivatives:

  • Outer derivative: $\cos(u)$
  • Inner derivative: $2x$

Step 3: Apply chain rule: $$\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x$$

$$\boxed{2x\cos(x^2)}$$


Example 4: Exponential with Chain Rule

Problem: Find $\frac{d}{dx}[e^{3x}]$

Step 1: Identify the parts:

  • Outer: $e^u$
  • Inner: $u = 3x$

Step 2: Find derivatives:

  • Outer derivative: $e^u$ (exponential is its own derivative!)
  • Inner derivative: $3$

Step 3: Apply chain rule: $$\frac{d}{dx}[e^{3x}] = e^{3x} \cdot 3$$

$$\boxed{3e^{3x}}$$


Example 5: Square Root (Fractional Power)

Problem: Find $\frac{d}{dx}[\sqrt{x^2+1}]$

Step 1: Rewrite as a power: $$\sqrt{x^2+1} = (x^2+1)^{1/2}$$

Step 2: Identify the parts:

  • Outer: $u^{1/2}$
  • Inner: $u = x^2 + 1$

Step 3: Find derivatives:

  • Outer derivative: $\frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}$
  • Inner derivative: $2x$

Step 4: Apply chain rule: $$\frac{d}{dx}[\sqrt{x^2+1}] = \frac{1}{2\sqrt{x^2+1}} \cdot 2x = \frac{2x}{2\sqrt{x^2+1}}$$

$$\boxed{\frac{x}{\sqrt{x^2+1}}}$$


Example 6: Double Chain Rule (Nested Functions)

Problem: Find $\frac{d}{dx}[\sin^2(3x)]$

Step 1: Recognize the structure — this is $[\sin(3x)]^2$, which has THREE layers:

  • Outermost: $(\cdot)^2$
  • Middle: $\sin(\cdot)$
  • Innermost: $3x$

Step 2: Work from outside in:

First, derivative of outer $u^2$ is $2u$, where $u = \sin(3x)$: $$2\sin(3x) \cdot \frac{d}{dx}[\sin(3x)]$$

Then, $\frac{d}{dx}[\sin(3x)] = \cos(3x) \cdot 3$

Step 3: Combine: $$\frac{d}{dx}[\sin^2(3x)] = 2\sin(3x) \cdot \cos(3x) \cdot 3$$

$$\boxed{6\sin(3x)\cos(3x)}$$

Or using the double angle identity: $\boxed{3\sin(6x)}$


Quick Reference: Common Chain Rule Patterns

Function Derivative
$[f(x)]^n$ $n[f(x)]^{n-1} \cdot f'(x)$
$\sin(f(x))$ $\cos(f(x)) \cdot f'(x)$
$\cos(f(x))$ $-\sin(f(x)) \cdot f'(x)$
$e^{f(x)}$ $e^{f(x)} \cdot f'(x)$
$\ln(f(x))$ $\frac{f'(x)}{f(x)}$
$\sqrt{f(x)}$ $\frac{f'(x)}{2\sqrt{f(x)}}$

Common Mistakes and Misunderstandings

❌ Mistake: Forgetting the chain rule entirely

Wrong: $\frac{d}{dx}[(2x+1)^4] = 4(2x+1)^3$

Why it's wrong: You used the power rule on the outer function but forgot to multiply by the derivative of the inner function ($2x+1$).

Correct: $\frac{d}{dx}[(2x+1)^4] = 4(2x+1)^3 \cdot 2 = 8(2x+1)^3$


❌ Mistake: Changing the inner function when differentiating

Wrong: $\frac{d}{dx}[\sin(x^2)] = \cos(2x) \cdot 2x$

Why it's wrong: The inner function $x^2$ should stay as $x^2$ inside the cosine, not become $2x$.

Correct: $\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x$


❌ Mistake: Applying chain rule when it's not needed

Wrong: $\frac{d}{dx}[x^3] = 3x^2 \cdot 1 = 3x^2$ (overthinking it)

Why it's unnecessary: When the inner function is just $x$, its derivative is 1, so the chain rule reduces to the basic rule. You don't need to explicitly write "$\cdot 1$".

Simpler: $\frac{d}{dx}[x^3] = 3x^2$ (just use the power rule directly)

Formulas & Reference

Chain Rule (Standard Form)

$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$

Derivative of a composite function: take the derivative of the outer function (keeping the inner function unchanged), then multiply by the derivative of the inner function

Variables:
$f$:
outer function
$g(x)$:
inner function
$f'(g(x))$:
derivative of outer, evaluated at the inner
$g'(x)$:
derivative of the inner function

Chain Rule (Leibniz Notation)

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

If y depends on u and u depends on x, the derivative of y with respect to x is the product of the two intermediate derivatives

Variables:
$y$:
dependent variable (function of u)
$u$:
intermediate variable (function of x)
$x$:
independent variable

Generalized Power Rule

$$\frac{d}{dx}[f(x)]^n = n[f(x)]^{n-1} \cdot f'(x)$$

Chain rule applied to a function raised to a power — most common use case!

Variables:
$f(x)$:
the inner function (base)
$n$:
the exponent (any real number)
$f'(x)$:
derivative of the inner function
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