Chain Rule with Product Rule examples

Hey there! 👋 It's super common to feel a bit tangled when both the Chain Rule and the Product Rule pop up in the same differentiation problem. You're definitely not alone! The trick is knowing which rule applies to the "outermost" structure of your function. Let's break it down with some clear examples!

Quick Recap: The Rules

• Product Rule: If you have a function $y = u(x)v(x)$, its derivative is $y' = u'(x)v(x) + u(x)v'(x)$.
• Chain Rule: If you have a composite function $y = f(g(x))$, its derivative is $y' = f'(g(x))g'(x)$. Think of it as "derivative of the outside, leaving the inside alone, multiplied by the derivative of the inside."

The Key Question: Which Rule Comes First? 🤔

This is where most people get stuck! You need to look at the overall structure of your function:

Is the entire function a product of two distinct terms, where *each* term might be a composite?
➡️ Apply Product Rule FIRST.

Is the entire function a composite function, where its "inner" part *is* a product?
➡️ Apply Chain Rule FIRST.

Example 1: Product Rule First (with Chain Rule inside)

Let's find the derivative of $y = x^2 \\cos(3x^2 + 1)$.

Here, we clearly have a product of two functions: $u(x) = x^2$ and $v(x) = \\cos(3x^2 + 1)$. So, the Product Rule is our first step.

1. Identify $u$ and $v$:
   • $u = x^2$
   • $v = \\cos(3x^2 + 1)$
2. Find $u'$ and $v'$:
   • $u' = 2x$
   • For $v'$, we need the Chain Rule!
     • Outer function: $\\cos(w)$, derivative is $-\\sin(w)$
     • Inner function: $w = 3x^2 + 1$, derivative is $w' = 6x$

     So, $v' = -\\sin(3x^2 + 1) \\cdot (6x) = -6x\\sin(3x^2 + 1)$.

3. Apply the Product Rule formula ($u'v + uv'$):

   $y' = (2x)(\\cos(3x^2 + 1)) + (x^2)(-6x\\sin(3x^2 + 1))$
4. Simplify:

   $y' = 2x\\cos(3x^2 + 1) - 6x^3\\sin(3x^2 + 1)$ ✅

Example 2: Chain Rule First (with Product Rule inside)

Now, let's differentiate $y = (x^3 \\sin(2x))^5$.

In this case, the outermost operation is raising something to the power of 5. The "something" inside is a product. Therefore, the Chain Rule is applied first.

1. Identify the "outer" and "inner" functions:
   • Outer function: $f(w) = w^5$, derivative is $f'(w) = 5w^4$.
   • Inner function: $w = x^3 \\sin(2x)$.
2. Apply the Chain Rule's first step ($f'(w)$):

   The derivative starts as $5(x^3 \\sin(2x))^4 \\cdot \\frac{d}{dx}(x^3 \\sin(2x))$.

3. Now, find the derivative of the inner function (\\frac{d}{dx}(x^3 \\sin(2x))$) using the Product Rule:
   • Let $a = x^3$ and $b = \\sin(2x)$.
   • $a' = 3x^2$.
   • For $b'$, we need the Chain Rule again!
     • Outer: $\\sin(z)$, derivative is $\\cos(z)$
     • Inner: $z = 2x$, derivative is $z' = 2$

     So, $b' = \\cos(2x) \\cdot (2) = 2\\cos(2x)$.
   • Apply Product Rule for $w'$ ($a'b + ab'$):

     \\frac{d}{dx}(x^3 \\sin(2x)) = (3x^2)(\\sin(2x)) + (x^3)(2\\cos(2x))$

     \\frac{d}{dx}(x^3 \\sin(2x)) = 3x^2\\sin(2x) + 2x^3\\cos(2x)$.

4. Combine everything for $y'$:

   $y' = 5(x^3 \\sin(2x))^4 [3x^2\\sin(2x) + 2x^3\\cos(2x)]$ ✨

You got this! Keep practicing, and always ask yourself "What's the main operation happening?" before diving into the individual parts. Good luck! 👍