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# Chain Rule

## Chain Rule Lesson

### What is the Chain Rule?

The chain rule is a method for differentiating composite functions. As a side note, we can think of composite functions as functions that contain other functions. So, in other words, the chain rule makes it easier for us to take the derivative of a function composed of other functions.

### Why do we Learn About the Chain Rule?

When we are in our calculus classes, the Chain Rule is extremely helpful when working with composite functions. Having the ability to take the derivative of a function composed of other functions is valuable, especially when it saves us from having to expand functions like (x2-3x+10)9 before taking the derivative.

But how can the Chain Rule help us when we are not in the classroom? To answer this question, let’s consider power consumption in our home at specific times, and in turn, using this information to evaluate the infrastructure requirements for a solar panel array or a generator.

First, a little background information on energy and power. When reading the electrical meter for a home, we see that we are given the unit of kWh, or kilowatt-hour. A kWh is a unit of energy (energy is the ability to do work). So, if we were to run a single 1000 W microwave oven for 1 hour, we would consume 1 kWh of energy (1,000 W = 1 kW). A watt, commonly notated with a “W”, is a unit of measurement for power. Power is the amount of energy consumed per unit time. If we consumed 1000 Joules of energy per second, we would have 1,000 W (1 kW) of power consumption.

Now that we have a better understanding of energy and power, let’s go back to our example. Let’s say that we are interested in installing auxiliary power sources for our home such as a solar panel array or a generator. If we were to record the average energy usage of our home (in kWh) for each hour of a day, we would likely get a composite function characterizing energy consumption with respect to time.

If we were to take the derivative of this function, we would get a function of power with respect to time. Plugging a specific value for the time variable into the function for power would then yield the power consumption of our home during a specified hour. This is equivalent to finding the slope of the energy vs. time curve at a specific time of the day.

The resulting information would enable us to evaluate our power requirements during peak hours of the day, which in turn, gives us insight into the infrastructure requirements for our new power source!

### How to use the Chain Rule

If we are given a function f(g(x)) that is differentiable at x, the derivative can be found using the following equation:

\begin{align}&\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\end{align}

Where f(g(x)) is the composite function, g(x) is a function that is differentiable at x, f'(g(x)) is the derivative of the outer portion of the composite function with respect to x, and g'(x) is the derivative of g(x) with respect to x.

The steps for taking the derivative of a composite function using the Chain Rule are as follows:

1. Identify the inner and outer functions. The inner function is g(x) and the outer function is f.
2. Take the derivative of the inner function g(x) to get g'(x).
3. Take the derivative of the outer function while leaving the inside function g(x) intact to get f'(g(x))
4. Multiply the result from Step 2 with the result from Step 3 to get f'(g(x))g'(x).
5. Simplify the result.

To see these steps in use, see the Chain Rule example problems below.

### Example Problem 1

\begin{align}& \text{Using the Chain Rule, find the derivative } \frac{d}{dx} \text{ of the given function:} \\ \\ & \hspace{3ex} 3(2x^{2}-5x-25)^{3}\\ \\ & \text{1.) The derivative of a composite function using the Chain Rule can be} \\ & \hspace{3ex} \text{found using the following equation:} \\ \\ & \hspace{3ex} \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \\ \\ & \hspace{3ex} \text{Where: } \\ & \hspace{3ex} f(g(x)) \text{ is the composite function, } g(x) \text{ is a function that is differentiable}\\ & \hspace{3ex} \text{at } x \text{, } f'(g(x)) \text{ is the derivative of the outer portion of the composite} \\ & \hspace{3ex} \text{function with respect to } x \text{, and } g'(x) \text{ is the derivative of } g(x) \text{ with respect} \\ & \hspace{3ex} \text{to } x \text{.} \\ \\ & \text{2.) Let’s begin by identifying the inner and outer functions of our given} \\ & \hspace{3ex} \text{composite function } f(g(x)) = 3(2x^{2}-5x-25)^{3} \text{ :} \\ \\ & \hspace{3ex} \text{2.1 ) } 3(\boxed{2x^{2}-5x-25})^{3} \\ \\ & \hspace{8ex} \Longrightarrow \text{ Therefore, the inner function } g(x) \text{ is } 2x^{2}-5x-25 \text{.} \\ \\ & \hspace{3ex} \text{2.2 ) } \boxed{3(}2x^{2}-5x-25\boxed{)^{3}} \\ \\ & \hspace{8ex} \Longrightarrow \text{Therefore, the outer function } f \text{ is } 3(g(x))^{3} \text{. In other words,}\\ & \hspace{13ex} \text{everything outside of the inner function } g(x) \text{ is considered} \\ & \hspace{13ex} \text{the “outside”.} \\ \\ & \text{3.) Next, let’s take the derivative of the inner function } g(x) \text{ to get } g'(x) \text{:} \\ \\ & \hspace{3ex} \Longrightarrow \text{From Step 2.1, } g(x) = 2x^{2}-5x-25 \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[g(x)] = 4x-5 \\ \\ & \hspace{3ex} \Longrightarrow \boxed{g'(x) = 4x-5} \\ \\ & \text{4.) Next, let’s take the derivative of the outer function } f \text{ to get } f'(g(x)) \text{:} \\ \\ & \hspace{3ex} \Longrightarrow \text{From Step 2.2, } f = 3(g(x))^{3} \\ \\ & \hspace{3ex} \Longrightarrow \frac{df}{dx} = 9(g(x))^{2} \\ \\ & \hspace{3ex} \Longrightarrow \boxed{f'(g(x)) = 9(2x^{2}-5x-25)^{2}} \\ \\ & \hspace{8ex} \text{Note: “} 2x^{2}-5x-25 \text{” was inserted in place of } g(x) \text{ since } \\ & \hspace{15ex} g(x) = 2x^{2}-5x-25 \text{ (from Step 2.1).} \\ \\ & \text{5.) Next, let’s multiply the result from Step 3 with the result from} \\ & \hspace{3ex} \text{Step 4 to obtain the derivative of our given composite function.} \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = 9(2x^{2}-5x-25)^{2} \cdot (4x-5) \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = (36x-45)(2x^{2}-5x-25)^{2} \\ \\ & \text{Therefore, the derivative } \frac{d}{dx} \text{ of the given composite function is:} \\ \\ & \boxed{\boxed{(36x-45)(2x^{2}-5x-25)^{2}}} \end{align}

### Example Problem 2

\begin{align}& \text{Using the Chain Rule, find the derivative } \frac{d}{dx} \text{ of the given function:} \\ \\ & \hspace{3ex} cos^{4}x \\ \\ & \text{1.) The derivative of a composite function using the Chain Rule can be} \\ & \hspace{3ex} \text{found using the following equation:} \\ \\ & \hspace{3ex} \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \\ \\ & \hspace{3ex} \text{Where: } \\ & \hspace{3ex} f(g(x)) \text{ is the composite function, } g(x) \text{ is a function that is differentiable}\\ & \hspace{3ex} \text{at } x \text{, } f'(g(x)) \text{ is the derivative of the outer portion of the composite} \\ & \hspace{3ex} \text{function with respect to } x \text{, and } g'(x) \text{ is the derivative of } g(x) \text{ with respect} \\ & \hspace{3ex} \text{to } x \text{.} \\ \\ & \text{2.) Let’s begin by identifying the inner and outer functions of our given} \\ & \hspace{3ex} \text{composite function } f(g(x)) = cos^{4}x = (cos \: x)^{4} \text{ :} \\ \\ & \hspace{3ex} \text{2.1 ) } (\boxed{cos\:x})^{4} \\ \\ & \hspace{8ex} \Longrightarrow \text{ Therefore, the inner function } g(x) \text{ is } cos\:x \text{.} \\ \\ & \hspace{3ex} \text{2.2 ) } \boxed{(}cos\:x\boxed{)^{4}} \\ \\ & \hspace{8ex} \Longrightarrow \text{Therefore, the outer function } f \text{ is } (g(x))^{4} \text{. In other words,}\\ & \hspace{13ex} \text{everything outside of the inner function } g(x) \text{ is considered} \\ & \hspace{13ex} \text{the “outside”.} \\ \\ & \text{3.) Next, let’s take the derivative of the inner function } g(x) \text{ to get } g'(x) \text{:} \\ \\ & \hspace{3ex} \Longrightarrow \text{From Step 2.1, } g(x) = cos\:x \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[g(x)] = -sin\:x \\ \\ & \hspace{3ex} \Longrightarrow \boxed{g'(x) = -sin\:x} \\ \\ & \text{4.) Next, let’s take the derivative of the outer function } f \text{ to get } f'(g(x)) \text{:} \\ \\ & \hspace{3ex} \Longrightarrow \text{From Step 2.2, } f = (g(x))^{4} \\ \\ & \hspace{3ex} \Longrightarrow \frac{df}{dx} = 4(g(x))^{3} \\ \\ & \hspace{3ex} \Longrightarrow \boxed{f'(g(x)) = 4(cos\:x)^{3}} \\ \\ & \hspace{8ex} \text{Note: “} cos\:x \text{” was inserted in place of } g(x) \text{ since } \\ & \hspace{15ex} g(x) = cos\:x \text{ (from Step 2.1).} \\ \\ & \text{5.) Next, let’s multiply the result from Step 3 with the result from} \\ & \hspace{3ex} \text{Step 4 to obtain the derivative of our given composite function.} \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = 4(cos\:x)^{3} \cdot (-sin\:x) \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = -4sin\:x(cos\:x)^{3} \\ \\ & \text{Therefore, the derivative } \frac{d}{dx} \text{ of the given composite function is:} \\ \\ & \boxed{\boxed{-4sin\:x(cos\:x)^{3}}} \end{align}

### Example Problem 3

\begin{align}& \text{Using the Chain Rule, find the derivative } \frac{d}{dx} \text{ of the given function:} \\ \\ & \hspace{3ex} cos\:x^{4} \\ \\ & \text{1.) The derivative of a composite function using the Chain Rule can be} \\ & \hspace{3ex} \text{found using the following equation:} \\ \\ & \hspace{3ex} \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \\ \\ & \hspace{3ex} \text{Where: } \\ & \hspace{3ex} f(g(x)) \text{ is the composite function, } g(x) \text{ is a function that is differentiable}\\ & \hspace{3ex} \text{at } x \text{, } f'(g(x)) \text{ is the derivative of the outer portion of the composite} \\ & \hspace{3ex} \text{function with respect to } x \text{, and } g'(x) \text{ is the derivative of } g(x) \text{ with respect} \\ & \hspace{3ex} \text{to } x \text{.} \\ \\ & \text{2.) Let’s begin by identifying the inner and outer functions of our given} \\ & \hspace{3ex} \text{composite function } f(g(x)) = cos\:x^{4} = cos \: (x^{4}) \text{ :} \\ \\ & \hspace{3ex} \text{2.1 ) } cos(\boxed{x^{4}}) \\ \\ & \hspace{8ex} \Longrightarrow \text{ Therefore, the inner function } g(x) \text{ is } x^{4} \text{.} \\ \\ & \hspace{3ex} \text{2.2 ) } \boxed{cos(}x^{4}\boxed{)} \\ \\ & \hspace{8ex} \Longrightarrow \text{Therefore, the outer function } f \text{ is } cos(g(x)) \text{. In other words,}\\ & \hspace{13ex} \text{everything outside of the inner function } g(x) \text{ is considered} \\ & \hspace{13ex} \text{the “outside”.} \\ \\ & \text{3.) Next, let’s take the derivative of the inner function } g(x) \text{ to get } g'(x) \text{:} \\ \\ & \hspace{3ex} \Longrightarrow \text{From Step 2.1, } g(x) = x^{4} \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[g(x)] = 4x^{3} \\ \\ & \hspace{3ex} \Longrightarrow \boxed{g'(x) = 4x^{3}} \\ \\ & \text{4.) Next, let’s take the derivative of the outer function } f \text{ to get } f'(g(x)) \text{:} \\ \\ & \hspace{3ex} \Longrightarrow \text{From Step 2.2, } f = cos(g(x)) \\ \\ & \hspace{3ex} \Longrightarrow \frac{df}{dx} = -sin(g(x)) \\ \\ & \hspace{3ex} \Longrightarrow \boxed{f'(g(x)) = -sin(x^{4})} \\ \\ & \hspace{8ex} \text{Note: “} x^{4} \text{” was inserted in place of } g(x) \text{ since } \\ & \hspace{15ex} g(x) = x^{4} \text{ (from Step 2.1).} \\ \\ & \text{5.) Next, let’s multiply the result from Step 3 with the result from} \\ & \hspace{3ex} \text{Step 4 to obtain the derivative of our given composite function.} \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = -sin(x^{4}) \cdot (4x^{3}) \\ \\ & \hspace{3ex} \Longrightarrow \frac{d}{dx}[f(g(x))] = -4x^{3}sin(x^{4}) \\ \\ & \text{Therefore, the derivative } \frac{d}{dx} \text{ of the given composite function is:} \\ \\ & \boxed{\boxed{-4x^{3}sin(x^{4})}} \end{align}
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