The **Chain Rule** is one of the most common derivatives applied in Differential Calculus (*or Calculus I*). It is used in deriving a composition of functions. The chain rule can be proven using the backbone of Calculus, which is the limits.

In this article, we will discuss everything about the chain rule. We will cover its definition, formula, proofs, and application usage. We will also look at some examples and practice problems to apply the principles of the chain rule.

## The Chain Rule and its Formula

### What is the Chain Rule?

The Chain Rule is defined as the derivative of a composition of at least two different types of function such as but not limited to:

$$y’ = \frac{d}{dx}[f \left( g(x) \right)]$$

where *g(x) *is a domain of function *f(u)*.

We can also call the function *f* as the outer function and function *g* as the inner function. In this composition, *f(x) and g(x)* must be two different types of functions that cannot be easily or possibly algebraically evaluated into a single type of function.

Remember that a composition of functions can be considered as a function within another function or a function of another function.

### The Chain Rule Formula

The Chain Rule formula can be verbally expressed as the derivative of the outer function *f* multiplied by the derivative of the inner function *g*. The inner function *g* is the domain of the derivative of the outer function *f*.

The chain rule formula can be illustrated as:

$$\frac{d}{dx} (f(g(x))) = \frac{d}{dx} \left(f(g(x)) \right) \cdot \frac{d}{x}(g(x))$$ |

where you derive *f(g(x))* by using the function *f’*s derivative method and using *g(x)* as the domain of the function *f * and then multiplying the derivative of function *f* by the derivative of *g(x)*.

In another form, it can also be illustrated as:

$$\frac{d}{dx} \left(f(g(x)) \right) = \frac{d}{du}(f(u)) \cdot \frac{d}{dx}(g(x))$$ |

where

- $latex f(u) =$ the outer function
- $latex u = g(x)$, the domain of the outer function $latex f(u)$
- $latex \frac{d}{du}(f(u)) =$ the derivative of the outer function $latex f(u)$ in terms of $latex u$
- $latex \frac{d}{dx}(g(x)) =$ the derivative of the inner function $latex g(x)$ in terms of $latex x$

## The difference between Chain Rule and Power Rule

There is a common misconception that the chain rule formula is an expanded form of the power rule formula or the power rule formula is a simpler form of the chain rule formula. But the truth is, these two formulas are different and both of them were proven differently.

A simple power function like $latex x^2$ can be derived using the chain rule formula if we consider $latex f(u) = u^2$ as an outer function of $latex f(g(x))$ and $latex g(x) = x$ as an inner function within the outer function *f. *

However, if you analyze it, we cannot derive the outer function *f,* which is a power function, without knowing the application of the power rule.

Therefore, in this formula

$$\frac{d}{dx}(u^n) = n \cdot u^{n-1} \cdot u’$$

where $latex n$ is an exponent and $latex u$ is any type of function being raised to an exponent $latex n$, we can consider this as one of the special cases or forms of the chain rule formula. But it does not mean that the chain rule and power rule are interchangeable.

In this context, we can say that this is an application of the power rule formula multiplied by a derivative of u, which is caused by the chain rule. This happens because the outer function is a power function whilst the inner function represented by $latex u$ can be any other type of function raised to an exponent $latex n$.

To expound this further, we have

$latex u^n = f(u)$

and

$latex u = g(x)$

where

- $latex f(u)$ is the outer function of a composite function
- $latex g(x)$ is the inner function within an outer function
*f*of a composite function - $latex u = g(x)$ is the domain of the outer function
*f*

Since this gives us a composition of two functions, it uses the chain rule formula to be derived. Without the power rule, it would be impossible to get the derivative of f'(u).

Since the principles of the power rule and chain rule are different and independent from each other, then we can state that these two rules are not interchangeable.

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## Proof of The Chain Rule

### Proof of The Chain Rule by using Limits

We know in Calculus that limits are considered the backbone of this mathematical branch. Therefore, it makes sense that most, if not all, derivative methods were founded using limits, and the chain rule is one of them.

By getting the derivatives using limits, (also known as getting the slope of a tangent line), we can simplify a formula that would satisfy the conditions stated by the chain rule.

By getting the limits of composite function $latex f(g(x))$ as *h* approaches *zero*,

$$\frac{d}{dx}(f(g(x))) = \lim \limits_{h \to 0} {\frac{f(g(x+h)) – f(g(x))}{h}}$$

we would arrive with the standard chain rule formula.

$$\frac{d}{dx} (f(g(x))) = \frac{d}{dx} \left(f(g(x)) \right) \cdot \frac{d}{x}(g(x))$$

To learn more about the proof of the chain rule using limits, please visit our article about Proof of The Chain Rule.

## When to use the Chain Rule to find derivatives

The chain rule formula is an efficient tool to use in order to derive composite functions like the following:

a. $latex H(x) = f(g(x))$, where g(x) is an inner function within the outer function *f*.

b. $latex H(x) = f \circ g(x)$, similar to *letter a* but written in alternative format.

c. $latex H(x) = f(g(h(j(x)))$, where *f* is the outermost function of a composition of four functions.

d. $latex H(x) = (f_{1…n} (x))$, where the composite *H(x)* may include a composition of multiple functions and $latex n$ denotes the total quantity of the functions in the composition.

These are the most common examples of functions that use the chain rule for derivation problems. Although, you may argue that a function can be algebraically operated or simplified before it can be derived using the initial and simpler derivative methods (or even a transcendental function derivative), that is not always the case.

Thus, we have the chain rule to make it still possible to derive a composition of functions that are either very difficult to algebraically operate or even impossible to be operated at all, especially in the case of transcendental functions.

## How to use the Chain Rule, a step by step tutorial

We are asked to derive

$latex H(x) = \sin{(x^3)}$

As you can observe, this given function can be considered a composite function. Therefore, we can use the chain rule formula to derive this problem.

**Step 1:** List down the chain rule formula for reference:

$$\frac{d}{dx} (H(x)) = \frac{d}{dx} \left(f(g(x)) \right) \cdot \frac{d}{x}(g(x))$$

Please take note that you may use any form of the chain rule formula as long as you find it more efficient based on your preference or the given problem.

** Step 2:** Identify how many functions we have in the given composite function. In this example, we have two. By listing these two functions, we have

If $latex g(x) = u$, then

$latex f(g(x)) = f(u)$

$latex f(u) = \sin{(u)}$

∴ $latex f(u)$ is a trigonometric function and will use the derivative of trigonometric functions to be derived

$latex u = g(x) = x^3$

∴ $latex g(x)$ is a power function and will use the power formula to be derived

If $latex f(g(x)) = f(u)$, then

$$\frac{d}{dx} [f(g(x))] = \frac{d}{du} [f(u)]$$

**Note:** We applied the substitution technique in deriving the outside function *f* to be more efficient and less confusing, especially for beginners.

** Step 3:** Let us now apply the chain rule formula:

$$\frac{d}{dx} H(x) = \frac{d}{du}(f(u)) \cdot \frac{d}{dx}(g(x))$$

$$\frac{d}{dx} H(x) = \frac{d}{du}(\sin{(u)}) \cdot \frac{d}{dx}(x^3)$$

$$\frac{d}{dx} H(x) = (\cos{(u)}) \cdot (3x^2)$$

** Step 4:** Substitute the inner function $latex g(x)$ into the $latex u$ of the derived equation:

$$\frac{d}{dx} H(x) = (\cos{(x^3)}) \cdot (3x^2)$$

** Step 5:** Simplify algebraically and apply the necessary trigonometric identities or other function-specific properties whenever applicable:

$$\frac{d}{dx} H(x) = 3x^2 \cdot \cos{(x^3)}$$

** Step 6:** If you think the derived equation cannot be simplified any longer, declare it as your final answer:

$latex H'(x) = 3x^2 \cos{(x^3)}$

## Chain Rule – Examples with answers

Each of the following examples has its respective detailed solution. It is recommended that you try to solve the sample problems yourself before looking at the solution so that you can practice and fully master this topic.

**EXAMPLE 1**

Derive: $latex H(x) = (12x+6)^{24}$

##### Solution

** Step 1:** List down the chain rule formula for reference:

$$\frac{d}{dx} (H(x)) = \frac{d}{dx} \left(f(g(x)) \right) \cdot \frac{d}{x}(g(x))$$

** Step 2:** By listing the two functions, we have

If $latex g(x) = u$, then

$latex f(g(x)) = f(u)$

$latex f(u) = u^{24}$

∴ $latex f(u)$ is a power function and will use the power formula to be derived

$latex u = g(x) = 12x+6$

∴ $latex g(x)$ is a polynomial function and will use the sum/difference of derivatives to be derived

If $latex f(g(x)) = f(u)$, then

$$\frac{d}{dx} [f(g(x))] = \frac{d}{du} [f(u)]$$

** Step 3:** Let us now apply the chain rule formula:

$$\frac{d}{dx} H(x) = \frac{d}{du}(f(u)) \cdot \frac{d}{dx}(g(x))$$

$$\frac{d}{dx} H(x) = \frac{d}{du}(u^{24}) \cdot \frac{d}{dx}(12x+6)$$

$$\frac{d}{dx} H(x) = (24u^{23}) \cdot (12)$$

** Step 4:** Substitute the inner function $latex g(x)$ into the $latex u$ of the derived equation:

$$\frac{d}{dx} H(x) = (24(12x+6)^{23}) \cdot (12)$$

** Step 5:** Simplify algebraically:

$$\frac{d}{dx} H(x) = 288 \cdot (12x+6)^{23}$$

** Step 6:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$latex H'(x) = 288(12x+6)^{23}$

**EXAMPLE 2**

Find the derivative of $latex f(x) = \sqrt[12]{6x-3}$.

##### Solution

** Step 1:** List down the chain rule formula for reference:

$$\frac{d}{dx} (H(x)) = \frac{d}{dx} \left(f(g(x)) \right) \cdot \frac{d}{x}(g(x))$$

** Step 2:** If $latex g(x) = u$, then

$latex f(g(x)) = f(u)$

$latex f(u) = \sqrt[12]{u}$

$latex f(u) = u^{\frac{1}{12}}$

∴ $latex f(u)$ is a radical converted into a power function and will use the power formula to be derived

$latex u = g(x) = 6x-3$

∴ $latex g(x)$ is a polynomial function and will use the sum/difference of derivatives to be derived

If $latex f(g(x)) = f(u)$, then

$$\frac{d}{dx} [f(g(x))] = \frac{d}{du} [f(u)]$$

** Step 3:** Let us now apply the chain rule formula:

$$\frac{d}{dx} H(x) = \frac{d}{du}(f(u)) \cdot \frac{d}{dx}(g(x))$$

$$\frac{d}{dx} H(x) = \frac{d}{du}(u^{\frac{1}{12}}) \cdot \frac{d}{dx}(6x-3)$$

$$\frac{d}{dx} H(x) = \left(\frac{1}{12}u^{-\frac{11}{12}} \right) \cdot (6)$$

** Step 4:** Substitute the inner function $latex g(x)$ into the $latex u$ of the derived equation:

$$\frac{d}{dx} H(x) = \left(\frac{1}{12} \cdot (6x-3)^{-\frac{11}{12}} \right) \cdot (6)$$

** Step 5:** Simplify algebraically:

$$\frac{d}{dx} H(x) = \frac{6}{12 \cdot (6x-3)^{\frac{11}{12}}}$$

$$\frac{d}{dx} H(x) = \frac{1}{2 \cdot (6x-3)^{\frac{11}{12}}}$$

** Step 6:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$$H'(x) = \frac{1}{2 \sqrt[12]{(6x-3)^{11}}}$$*in radical form*

**EXAMPLE 3**

Derive: $latex \cos{(12x^2+6x-3)}$.

##### Solution

** Step 1:** List down the chain rule formula for reference:

$$\frac{d}{dx} (H(x)) = \frac{d}{dx} \left(f(g(x)) \right) \cdot \frac{d}{x}(g(x))$$

** Step 2:** In this example, we have $latex g(x) = u$, then

$latex f(g(x)) = f(u)$

$latex f(u) = \cos{(u)}$

∴ $latex f(u)$ is a trigonometric function and will use the derivative of trigonometric functions to be derived

$latex u = g(x) = 12x^2+6x-3$

∴ $latex g(x)$ is a polynomial function and will use the sum/difference of derivatives and power formula to be derived

If $latex f(g(x)) = f(u)$, then

$$\frac{d}{dx} [f(g(x))] = \frac{d}{du} [f(u)]$$

** Step 3:** Let us now apply the chain rule formula:

$$\frac{d}{dx} H(x) = \frac{d}{du}(f(u)) \cdot \frac{d}{dx}(g(x))$$

$$\frac{d}{dx} H(x) = \frac{d}{du}(\cos{(u)}) \cdot \frac{d}{dx}(12x^2+6x-3)$$

$$\frac{d}{dx} H(x) = (-\sin{(u)}) \cdot (24x+6)$$

** Step 4:** Substitute the inner function $latex g(x)$ into the $latex u$ of the derived equation:

$$\frac{d}{dx} H(x) = (-\sin{(12x^2+6x-3)}) \cdot (24x+6)$$

** Step 5:** Simplify algebraically:

$$\frac{d}{dx} H(x) = -(24+6) \cdot \sin{(12x^2+6x-3)}$$

** Step 6:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$$H'(x) = – (24 + 6) \sin{(12x^2+6x-3)}$$

**EXAMPLE 4**

What is the derivative of $latex \csc{\ln{(12x+6)}}$?

##### Solution

** Step 1:** List down the chain rule formula for reference:

$$\frac{d}{dx} (H(x)) = \frac{d}{dx} \left(f(g(x)) \right) \cdot \frac{d}{x}(g(x))$$

** Step 2:** Identify how many functions you have in the problem. In this example, there are three. By listing these three functions, we have

If $latex g(h(x)) = u$, then

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

$latex f(u) = \csc{(u)}$

∴ $latex f(u)$ is a trigonometric function and will use the derivative of trigonometric functions to be derived

If $latex g(h(x)) = v$, then

$latex g(h(x)) = g(v)$

$latex g(v) = \ln{(v)}$

∴ $latex g(v)$ is a logarithmic function and will use the derivative of logarithmic functions to be derived

$latex v = h(x) = 12x+6$

∴ $latex h(x)$ is a polynomial function and will use sum/difference of derivatives to be derived

If $latex f(g(h(x))) = f(u)$, then

$$\frac{d}{dx} [f(g(h(x)))] = \frac{d}{du} [f(u)]$$

If $latex g(h(x)) = g(v)$, then

$$\frac{d}{dx} [g(h(x))] = \frac{d}{dv} [g(v)]$$

** Step 3:** Let us now apply the chain rule formula:

$$f_{1…n}'(x) = f_1′ \left( f_{2…n}(x) \right) \cdot f_2′ \left( f_{3…n}(x) \right)\cdots f_{n-1}’ \left(f_{n…n}(x)\right) \cdot f_n'(x)$$

$$\frac{d}{dx} H(x) = \frac{d}{du} f(u) \cdot \frac{d}{dv} g(v) \cdot \frac{d}{dx} h(x)$$

$$\frac{d}{dx} H(x) = \frac{d}{du}(\csc{(u)}) \cdot \frac{d}{dv}(\ln{(v)}) \cdot \frac{d}{dx}(12x+6)$$

$$\frac{d}{dx} H(x) = (-\csc{(u)} \cot{(u)}) \cdot (\frac{1}{v}) \cdot {12}$$

** Step 4:** Substitute $latex g(h(x))$ and $latex h(x)$ into $latex u$ and $latex v$:

$$\frac{d}{dx} H(x) = (-\csc{(\ln{(12x+6)})} \cot{(\ln{(12x+6)})})\cdot (\frac{1}{12x+6}) \cdot {12}$$

** Step 5:** Simplify algebraically:

$$\frac{d}{dx} H(x) = \frac{-12 \csc{(\ln{(12x+6)})} \cot{(\ln{(12x+6)})}}{12x+6}$$

$$\frac{d}{dx} H(x) = \frac{-12 \csc{(\ln{(12x+6)})} \cot{(\ln{(12x+6)})}}{6(x+2)}$$

$$\frac{d}{dx} H(x) = \frac{-2 \csc{(\ln{(12x+6)})} \cot{(\ln{(12x+6)})}}{(x+2)}$$

** Step 6:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$$H'(x) = -\frac{2 \csc{(\ln{(12x+6)})} \cot{(\ln{(12x+6)})}}{(x+2)}$$

**EXAMPLE 5**

Derive: $latex e^{\sin^{2}{(6x-3)}}$

##### Solution

** Step 1:** List down the chain rule formula for reference:

$$\frac{d}{dx} (H(x)) = \frac{d}{dx} \left(f(g(x)) \right) \cdot \frac{d}{x}(g(x))$$

** Step 2:** Identify how many functions you have in the problem. In this example, there are four. By listing these four functions, we have

If $latex f(g(h(j(x)))) = u$, then

$latex f(g(h(j(x)))) = f(u)$

$latex f(u) = e^u$

∴ $latex f(u)$ is an exponential function and will use the derivative of exponential functions to be derived

If $latex g(h(j(x))) = v$, then

$latex g(h(j(x))) = g(v)$

$latex g(v) = v^2$

∴ $latex g(v)$ is a power function and will use the power rule to be derived

If $latex h(j(x)) = w$, then

$latex h(j(x)) = h(w)$

$latex h(w) = \sin{(w)}$

∴ $latex h(w)$ is a trigonometric function and will use the derivative of trigonometric functions to be derived

$latex w = j(x) = 6x-3$

∴ $latex j(x)$ is a polynomial function and will use sum/difference of derivatives to be derived

If $latex f(g(h(j(x)))) = f(u)$, then

$$\frac{d}{dx} [f(g(h(j(x))))] = \frac{d}{du} [f(u)]$$

If $latex g(h(j(x))) = g(v)$, then

$$\frac{d}{dx} [g(h(j(x)))] = \frac{d}{dv} [g(v)]$$

If $latex h(j(x)) = h(w)$, then

$$\frac{d}{dx} [h(j(x))] = \frac{d}{dw} [h(w)]$$

** Step 3:** Let us now apply the chain rule formula:

$$f_{1…n}'(x) = f_1′ \left( f_{2…n}(x) \right) \cdot f_2′ \left( f_{3…n}(x) \right)\cdots f_{n-1}’ \left(f_{n…n}(x)\right) \cdot f_n'(x)$$

$$\frac{d}{dx} H(x) = \frac{d}{du} f(u) \cdot \frac{d}{dv} g(v) \cdot \frac{d}{dw} h(w) \cdot \frac{d}{dx} j(x)$$

$$\frac{d}{dx} H(x) = \frac{d}{du} (e^u) \cdot \frac{d}{dv} (v^2)\cdot \frac{d}{dw} (\sin{(w)}) \cdot \frac{d}{dx} (6x-3)$$

$$\frac{d}{dx} H(x) = (e^u) \cdot (2v) \cdot (\cos{(w)}) \cdot (6)$$

** Step 4:** Substitute $latex g(h(j(x)))$, $latex h(j(x))$, and $latex j(x)$ into $latex u$, $latex v$, and $latex w$:

$$\frac{d}{dx} H(x) = (e^{\sin^{2}{(6x-3)}}) \cdot (2(\sin{(6x-3)}))\cdot (\cos{(6x-3)}) \cdot (6)$$

** Step 5:** Simplify algebraically:

$$\frac{d}{dx} H(x) = 12 \cdot \sin{(6x-3)} \cdot \cos{(6x-3)} \cdot e^{\sin^{2}{(6x-3)}}$$

** Step 6:** If you think the derived equation cannot be simplified any longer, declare it as your final derivative answer:

$$H'(x) = 12 \sin{(6x-3)} \cos{(6x-3)} e^{\sin^{2}{(6x-3)}}$$

## Chain Rule – Practice problems

Solve the following differentiation problems and test your knowledge on this topic. Use the chain rule formula detailed above to solve the exercises. If you have problems with these exercises, you can study the examples solved above.

## See also

Interested in learning more about the chain rule? Take a look at these pages:

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