The Chain Rule is an important tool in calculus that helps us find the derivative of composite functions. Knowing how to use the Chain Rule is crucial because you will see it often in more advanced math. In this lesson, we will go through what the Chain Rule is, how to use it, and provide some examples and practice problems.

What is the Chain Rule?

The Chain Rule is useful when we have a function made up of two or more functions combined together.

For example, think of a function like ( h(x) = f(g(x)) ).

Here, ( f ) is the outer function, and ( g ) is the inner function.

A common example is ( h(x) = \sin(x^2) ). In this case:

• The outer function is ( f(u) = \sin(u) )
• The inner function is ( g(x) = x^2 )

We use the Chain Rule when we need to find out how a small change in ( x ) affects the result of ( h(x) ).

To do this correctly, we need to understand how changes in ( g(x) ) affect ( f(g(x)) ).

How to Derive the Chain Rule Formula

To understand the Chain Rule better, let’s break down how to derive it step by step.

Starting with the function:

$h(x) = f(g(x))$

The Chain Rule tells us that:

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

Let’s see how we get to this formula.

1. We begin by using the limit definition of the derivative: $h'(x) = \lim_{h \to 0} \frac{h(x + h) - h(x)}{h}$

2. Plugging our function into this gives: $h'(x) = \lim_{h \to 0} \frac{f(g(x + h)) - f(g(x))}{h}$

3. As ( h ) gets very small, ( g(x + h) ) gets closer to ( g(x) ). To make this clearer, let’s introduce ( k ) where: $k = g(x + h) - g(x)$

4. So now we can rewrite the limit: $\frac{f(g(x + h)) - f(g(x))}{g(x + h) - g(x)} \cdot \frac{g(x + h) - g(x)}{h}$

5. Using the derivative definitions ( f'(g(x)) ) for the first part and ( g'(x) ) for the second brings us to: $h'(x) = f'(g(x)) \cdot g'(x)$

And that is the Chain Rule formula! It helps us efficiently differentiate composite functions.

Step-by-Step Examples of the Chain Rule

Let’s look at some examples to see how to apply the Chain Rule.

Example 1: Differentiate ( h(x) = \sin(x^2) )

1. Identify the outer and inner functions:

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

2. Find the derivatives:

   • ( f'(u) = \cos(u) )
   • ( g'(x) = 2x )

3. Use the Chain Rule: $h'(x) = f'(g(x)) \cdot g'(x) = \cos(g(x)) \cdot 2x$

4. Replace ( g(x) ): $h'(x) = \cos(x^2) \cdot 2x = 2x \cos(x^2)$

Example 2: Differentiate ( h(x) = e^{3x + 1} )

1. Identify the functions:

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

2. Find the derivatives:

   • ( f'(u) = e^u )
   • ( g'(x) = 3 )

3. Use the Chain Rule: $h'(x) = f'(g(x)) \cdot g'(x) = e^{g(x)} \cdot 3$

4. Replace ( g(x) ): $h'(x) = 3e^{3x + 1}$

Example 3: Differentiate ( h(x) = \ln(5x^2 + 3) )

1. Identify the functions:

   • Outer: ( f(u) = \ln(u) )
   • Inner: ( g(x) = 5x^2 + 3 )

2. Find the derivatives:

   • ( f'(u) = \frac{1}{u} )
   • ( g'(x) = 10x )

3. Use the Chain Rule: $h'(x) = f'(g(x)) \cdot g'(x) = \frac{1}{g(x)} \cdot 10x$

4. Replace ( g(x) ): $h'(x) = \frac{10x}{5x^2 + 3}$

Example 4: Differentiate ( h(x) = \sqrt{2x^3 + 1} )

1. Identify the functions:

   • Outer: ( f(u) = \sqrt{u} )
   • Inner: ( g(x) = 2x^3 + 1 )

2. Find the derivatives:

   • ( f'(u) = \frac{1}{2\sqrt{u}} )
   • ( g'(x) = 6x^2 )

3. Use the Chain Rule: $h'(x) = f'(g(x)) \cdot g'(x) = \frac{1}{2\sqrt{g(x)}} \cdot 6x^2$

4. Replace ( g(x) ): $h'(x) = \frac{6x^2}{2\sqrt{2x^3 + 1}} = \frac{3x^2}{\sqrt{2x^3 + 1}}$

Practice Problems

Now that we learned about the Chain Rule with examples, it's your turn! Here are some practice problems to try.

1. Differentiate:
   $h(x) = (3x + 2)^4$

   • Identify the outer and inner functions and find the derivative.

2. Differentiate:
   $h(x) = \tan(5x^2 + 1)$

   • Identify both functions and find the derivative.

3. Differentiate:
   $h(x) = \cos(x^3 - 3x)$

   • Identify both functions and find the derivative using the Chain Rule.

4. Differentiate:
   $h(x) = \frac{1}{\ln(5x)}$

   • Identify both functions and find the derivative.

5. Differentiate:
   $h(x) = e^{x^3 + 2}$

   • Identify both functions and find the derivative.

Conclusion

By understanding the Chain Rule, you now have a powerful tool to handle many composite functions in calculus. Remember, recognizing the outer and inner functions is critical for successfully applying the Chain Rule. The more you practice identifying these functions and finding their derivatives, the better you will become at this important concept in calculus!