Lesson Explainer: Combining the Product, Quotient, and Chain Rules | Nagwa Lesson Explainer: Combining the Product, Quotient, and Chain Rules | Nagwa

Lesson Explainer: Combining the Product, Quotient, and Chain Rules Mathematics

In this explainer, we will learn how to find the first derivative of a function using combinations of the product, quotient, and chain rules.

Many functions can be constructed from simpler functions by combining them in one or more of the following ways:

  • addition and subtraction: 𝑢(𝑥)+𝑣(𝑥) and 𝑢(𝑥)𝑣(𝑥),
  • multiplication and division: 𝑢(𝑥)𝑣(𝑥) and 𝑢(𝑥)𝑣(𝑥),
  • composition: 𝑢(𝑣(𝑥)).

Fortunately, we recall that there are rules for differentiating functions that are formed in these ways. For addition and subtraction, we can use the linearity of the derivative; for multiplication and division, we have the product rule and quotient rule; for composition, we can apply the chain rule. Let us review these rules.

Rule: Rules of Differentiation

For differentiable functions 𝑢(𝑥) and 𝑣(𝑥) and constants 𝑎,𝑏, we have the following rules:

  • Linearity: (𝑎𝑢(𝑥)+𝑏𝑣(𝑥))=𝑎𝑢(𝑥)+𝑏𝑣(𝑥).
  • Product rule: (𝑢(𝑥)𝑣(𝑥))=𝑢(𝑥)𝑣(𝑥)+𝑢(𝑥)𝑣(𝑥).
  • Quotient rule: for 𝑣(𝑥)0, 𝑢(𝑥)𝑣(𝑥)=𝑢(𝑥)𝑣(𝑥)𝑢(𝑥)𝑣(𝑥)(𝑣(𝑥)).
  • Chain rule: (𝑢(𝑣(𝑥)))=𝑢(𝑣(𝑥))𝑣(𝑥).

In addition to using these rules separately, it is also possible to use them in conjunction with each other, allowing us to differentiate any combination of elementary functions. However, we should be aware that this is often not a trivial exercise and it can be challenging to identify the correct rules to apply, the best order to apply them, and whether there are algebraic simplifications that will make the process easier. In this explainer, we will look at a number of examples that will highlight the skills we need to navigate this landscape.

Let us consider an example of differentiating a complicated function combining many different operations together, and how we can tackle the differentiation by splitting it into separate parts. Suppose we have 𝑓(𝑥)=𝑥+4𝑥3𝑥+6+2𝑥5.

At first, this may seem impossible to deal with, but we can break it into parts. Generally, the best way to do this is to begin by considering the outermost layer first and working inward. If we do this, we can see that it is the sum of two functions: 𝑓(𝑥)=𝑥+4𝑥3𝑥+6+2𝑥5=𝑔(𝑥)+(𝑥).()()

Using the linearity of differentiation, this means we can differentiate 𝑔(𝑥) and (𝑥) separately and add them together afterwards. If we consider 𝑔(𝑥) separately in this way, we can see that 𝑔(𝑥)=𝑥+4𝑥3𝑥+6=(𝑢(𝑥)).()

In other words, 𝑔(𝑥) is a composition of functions, so we can apply the chain rule to help us differentiate it. This of course means that we need to find the derivative of 𝑢(𝑥), but this too can be broken down into smaller parts. Continuing this process, we can continue to remove layers of complexity from the function until we arrive at elementary expressions that we know how to differentiate. We can represent this visually as follows.

Notice that all the functions at the bottom of the tree are functions that we can differentiate easily. Hence, we can see that by applying the appropriate rules at each stage, we can find the derivatives of even the most complicated functions.

We will now look at some examples where we apply this approach, albeit to slightly simpler cases. To start with, let us consider a high-degree polynomial function that has been factored.

Example 1: Finding the First Derivative of Polynomial Functions at a Point Using the Product and Chain Rules

Find the first derivative of 𝑦=(𝑥5)(𝑥2) at (1,4).


Let us first analyze the given function and see what rules we can apply to it. Since it is a polynomial function, it would be possible for us to expand all the parentheses via multiplication and take the derivatives of the separate components, but this would require a large number of calculations. Instead, it would be more efficient to split it into parts by combining a couple of rules of differentiation together. We first of all note that this is a product of functions: 𝑦=(𝑥5)(𝑥2)=𝑢(𝑥)𝑣(𝑥).()()

Recall that the product rule tells us that the derivative of the product of two differentiable functions is given by (𝑢(𝑥)𝑣(𝑥))=𝑢(𝑥)𝑣(𝑥)+𝑢(𝑥)𝑣(𝑥).

To use this formula, we need to know the derivatives 𝑢(𝑥) and 𝑣(𝑥). The first function, 𝑢(𝑥)=𝑥5, can be differentiated by using the power rule to get 𝑢(𝑥)=1. For 𝑣(𝑥), we note that it is a composition of functions: 𝑣(𝑥)=(𝑥2)=(𝑔(𝑥)).()

That is, if 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=𝑥2, then 𝑣(𝑥)=𝑓(𝑔(𝑥)). We recall that the derivative of a composition of functions can be found using the chain rule, which is (𝑓(𝑔(𝑥)))=𝑓(𝑔(𝑥))𝑔(𝑥).

Using the power rule, we find that the derivative of 𝑓 is 𝑓(𝑥)=6𝑥 and the derivative of 𝑔 is 𝑔(𝑥)=1. Substituting these into the formula above (along with 𝑔(𝑥)=𝑥2), we get 𝑣(𝑥)=𝑓(𝑥2)1=6(𝑥2).

Now that we have both 𝑣(𝑥) and 𝑢(𝑥), we can substitute 𝑢(𝑥)=𝑥5, 𝑣(𝑥)=(𝑥2), 𝑢(𝑥)=1, and 𝑣(𝑥)=6(𝑥2) into the product rule formula to get dd𝑦𝑥=𝑢(𝑥)𝑣(𝑥)+𝑢(𝑥)𝑣(𝑥)=1(𝑥2)+(𝑥5)6(𝑥2)=(𝑥2)((𝑥2)+6(𝑥5))=(𝑥2)(7𝑥32).

Finally, we need to find the first derivative at the point (1,4). Note that (1,4) is the coordinate of the point on the graph representing the given function, and at this point, the 𝑥-coordinate is equal to 1. Thus, we substitute in 𝑥=1 to find dd𝑦𝑥|||=(12)(7132)=(1)(25)=25.

As we can see, it is often useful to be able to break down functions into components that we can deal with individually by using the product and chain rules. Let us consider another example where it is beneficial for us to combine different rules together.

Example 2: Differentiating Combinations of Polynomial and Root Functions Using Product and Chain Rules

Find dd𝑥5𝑥2𝑥+2 at 𝑥=1.


We have been asked to differentiate a function that is a combination of polynomial and square root functions, so let us consider how best to deal with it. One possibility is to take the 5𝑥 term into the square root and use the chain rule to differentiate the resulting square root function. The other way is to recognize that it is a product of functions, as follows: 5𝑥2𝑥+2=𝑢(𝑥)𝑣(𝑥).()()

Thus, since 𝑢(𝑥) and 𝑣(𝑥) are differentiable functions, we can use the product rule, which says that (𝑢(𝑥)𝑣(𝑥))=𝑢(𝑥)𝑣(𝑥)+𝑢(𝑥)𝑣(𝑥).

To use this, we must calculate 𝑢(𝑥) and 𝑣(𝑥). First of all, 𝑢(𝑥)=5𝑥 can be differentiated using the power rule to get 𝑢(𝑥)=5. For 𝑣(𝑥), we cannot use the power rule directly, but we can break it down into parts by recognizing that it is a composition of functions, as follows: 𝑣(𝑥)=2𝑥+2=𝑔(𝑥).()

That is, 𝑣(𝑥)=𝑓(𝑔(𝑥)), where 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=2𝑥+2. Thus, we can use the chain rule, which is (𝑓(𝑔(𝑥)))=𝑓(𝑔(𝑥))𝑔(𝑥).

Now, we can find both 𝑓(𝑥) and 𝑔(𝑥) using the power rule, which gives us 𝑓(𝑥)=12𝑥,𝑔(𝑥)=4𝑥.

Substituting these into the chain rule equation, we get 𝑣(𝑥)=12𝑔(𝑥)4𝑥=4𝑥22𝑥+2=2𝑥2𝑥+2.

Taking this function for 𝑣(𝑥), along with 𝑢(𝑥)=5, 𝑢(𝑥)=5𝑥, and 𝑣(𝑥)=2𝑥+2, we can put them into the product rule equation to get dd𝑥5𝑥2𝑥+2=52𝑥+2+5𝑥2𝑥2𝑥+2=52𝑥+2+10𝑥2𝑥+2.

Now, recall that we need to evaluate this function at 𝑥=1. Doing this, we get the following solution: dd𝑥5𝑥2𝑥+2||=521+2+10121+2=54+104=10+5=15.

In the first couple of examples, we have considered functions that require both the product rule and the chain rule to be differentiated. In the next example, we will consider a function defined in terms of polynomials and square root functions where we will need to use the quotient rule.

Example 3: Differentiating a Function Involving Rational Functions at a Point Using the Rules of Differentiation

Evaluate dd𝑦𝑥 at (1,1) if 𝑦=2𝑥3𝑥+1.


In this example, we want to determine the first derivative of the given rational function by applying the rules of differentiation and evaluating this at the point (1,1).

Since we will most likely have to apply more than one rule of differentiation, it will be helpful to consider what order we want to apply them in. The best way to do this is to consider the outermost part of the function and see how it can be broken down; if we do this, we can see that the given function is a quotient of functions as follows: 𝑦=2𝑥3𝑥+1=𝑢(𝑥)𝑣(𝑥).()()

In order to determine the derivative of this, we will make use of the quotient rule which states that for a function 𝑦=𝑢(𝑥)𝑣(𝑥), we have, for 𝑣(𝑥)0, dd𝑦𝑥=𝑢(𝑥)𝑣(𝑥)𝑢(𝑥)𝑣(𝑥)𝑣(𝑥).

Thus, we can determine the derivative of the given function by setting 𝑢(𝑥)=2𝑥 and 𝑣(𝑥)=3𝑥+1 in the numerator and denominator respectively. In order to apply this rule, we need to evaluate the derivatives of both 𝑢(𝑥) and 𝑣(𝑥).

The derivative of 𝑢(𝑥) is straightforward and can be found from the power rule as 𝑢(𝑥)=2. Since 𝑣(𝑥) is the composition of two functions, 𝑓(𝑥)=𝑥 and 𝑔(𝑥)=3𝑥+1, to find the derivative of 𝑣(𝑥), we will make use of the chain rule which states that for a composite function 𝑓(𝑔(𝑥)), we have (𝑓(𝑔(𝑥)))=𝑓(𝑔(𝑥))𝑔(𝑥).

In order to apply this rule, we need to evaluate the derivatives of 𝑓 and 𝑔 with respect to 𝑥, both of which can be found by an application of the power rule as follows: 𝑓(𝑥)=12𝑥,𝑔(𝑥)=6𝑥.

Substituting these expressions back into the chain rule with 𝑔(𝑥)=3𝑥+1, we find the derivative of 𝑣 with respect to 𝑥 as follows: 𝑣(𝑥)=3𝑥3𝑥+1.

Substituting 𝑢=2𝑥, 𝑢(𝑥)=2, 𝑣(𝑥)=3𝑥+1, and 𝑣(𝑥) as shown above, we can now apply the quotient rule as follows: dd𝑦𝑥=𝑢(𝑥)𝑣(𝑥)𝑢(𝑥)𝑣(𝑥)𝑣(𝑥)=23𝑥+1+2𝑥3𝑥+1.

By multiplying both the numerator and denominator by 3𝑥+1, we can simplify the fraction, giving us dd𝑦𝑥=23𝑥+1+2𝑥3𝑥(3𝑥+1)3𝑥+1=6𝑥2+6𝑥(3𝑥+1)3𝑥+1=2(3𝑥+1)3𝑥+1.

Substituting the given point 𝑥=1 in this expression gives dd𝑦𝑥|||=2(3+1)3+1=14.

So far, we have only seen instances of examples where it is optimal to apply the product or quotient rule followed by the chain rule, but the opposite order may be more natural depending on the given function. In each case, we have to make sure that the order in which we are applying the derivative rules makes sense. Let us consider a situation where we may have to change the order of applying the rules.

Example 4: Differentiating a Composition of Rational and Root Functions Using the Chain and Quotient Rules

If 𝑦=2𝑥+12𝑥1, determine dd𝑦𝑥.


Here, we have been asked to calculate the derivative of a function that appears to be a combination of different functions. Since we will probably have to use more than one rule of differentiation to do this, we should begin by considering which one to apply first. We do this by looking at the outermost part of the function and working inward. Since the outer part is a square root, we can see that we have 𝑦=2𝑥+12𝑥1=𝑣(𝑥).()

That is, 𝑦=𝑢(𝑣(𝑥)), where 𝑢(𝑥)=𝑥 and 𝑣(𝑥)=2𝑥+12𝑥1. We note that since this is a composition of differentiable functions, we can use the chain rule, which says that (𝑢(𝑣(𝑥)))=𝑢(𝑣(𝑥))𝑣(𝑥).

To use this formula, we need 𝑢(𝑥) and 𝑣(𝑥), the former of which can just be obtained by using the power rule to get 𝑢(𝑥)=12𝑥. For the latter, we can see that it is a quotient of differentiable functions: 𝑣(𝑥)=2𝑥+12𝑥1=𝑓(𝑥)𝑔(𝑥).()()

Therefore, we can apply the quotient rule. Recall that for 𝑔(𝑥)0, this is 𝑓(𝑥)𝑔(𝑥)=𝑓(𝑥)𝑔(𝑥)𝑓(𝑥)𝑔(𝑥)(𝑔(𝑥)).

Each of 𝑓(𝑥) and 𝑔(𝑥) is a polynomial function, so we can differentiate them using the power rule. This gives us 𝑓(𝑥)=6𝑥,𝑔(𝑥)=6𝑥.

Then, substituting these functions into the quotient rule formula, we get 𝑣(𝑥)=𝑓(𝑥)𝑔(𝑥)𝑓(𝑥)𝑔(𝑥)(𝑔(𝑥))=6𝑥2𝑥12𝑥+16𝑥(2𝑥1)=12𝑥(2𝑥1).

Now that we have 𝑣(𝑥), and 𝑢(𝑥)=12𝑥, let us recall that our original derivative could be found using the chain rule as follows: dd𝑦𝑥=𝑢(𝑣(𝑥))𝑣(𝑥)=12(𝑣(𝑥))12𝑥(2𝑥1).

Recall that 𝑣(𝑥)=2𝑥+12𝑥1, which means that 𝑣(𝑥)=2𝑥12𝑥+1. If we substitute this into the above equation, we get dd𝑦𝑥=122𝑥12𝑥+112𝑥(2𝑥1).

We can simplify this a bit by making use of the fact that 𝑎𝑏=𝑎𝑏 to get dd𝑦𝑥=122𝑥12𝑥+112𝑥(2𝑥1)=6𝑥2𝑥12𝑥+1(2𝑥1)2=6𝑥2𝑥+1(2𝑥1).

We note that there is an alternate way to solve the above problem. If we had used the fact that 𝑎𝑏=𝑎𝑏 from the start, we would have 𝑦=2𝑥+12𝑥1.

If we took the derivative of this, we could use the quotient rule first instead, followed by the chain rule on the functions in the numerator and denominator. However, this would result in a similar amount of work to get to the same answer.

Recall that there exists another form of differentiating problem that we might encounter. Suppose we have to find dd𝑦𝑥, where 𝑦=𝑓(𝑧),𝑧=𝑔(𝑥).

If 𝑦 is not given explicitly in terms of 𝑥, we cannot differentiate it directly. Instead, we can use the following form of the chain rule: dddddd𝑦𝑥=𝑦𝑧𝑧𝑥.

Since we can differentiate 𝑦 with respect to 𝑧 and 𝑧 with respect to 𝑥, this allows us to compute the derivative. Alternatively, we could substitute 𝑧=𝑔(𝑥) into the equation for 𝑦 to get 𝑦=𝑓(𝑔(𝑥)).

Recall that, written this way, we can return to the original form of the chain rule that we have been using so far: (𝑓(𝑔(𝑥)))=𝑓(𝑔(𝑥))𝑔(𝑥).

We note that this is just another way of writing dd𝑦𝑥 as shown above (in particular, 𝑓(𝑔(𝑥))=𝑦𝑧dd and 𝑔(𝑥)=𝑧𝑥dd). It is also important to note that substituting 𝑧=𝑔(𝑥) into the equation for 𝑦 in this way can lead to the expression simplifying, meaning there is a possibility we could differentiate it without using the chain rule (but this is not always the case).

In any case, let us consider an example in this style where we have to use the chain rule, with the additional complication of having to use another rule of differentiation.

Example 5: Differentiating a Composition of Rational Functions Using the Chain and Quotient Rules

Evaluate dd𝑦𝑥 at 𝑥=4 if 𝑦=𝑧+3𝑧+13 and 𝑧=𝑥10𝑥3.


In this example, we need to differentiate a function with respect to 𝑥 that is not explicitly given in terms of 𝑥.

One possibility of differentiating this would be to substitute the equation for 𝑧 into the function for 𝑦, so that we can differentiate 𝑦 directly with respect to 𝑥. However, this would result in fractions inside of fractions, which might be messy.

Instead, let us use the chain rule to differentiate this. Recall that the chain rule can be written in terms of d𝑦’s and d𝑥’s (i.e., Leibniz’s notation) as follows: dddddd𝑦𝑥=𝑦𝑧𝑧𝑥.

Thus, let us compute dd𝑦𝑧 and dd𝑧𝑥. Both functions are quotients of polynomials (which are differentiable functions), which means we can use the quotient rule to differentiate them. Recall that for 𝑣(𝑥)0, the quotient rule is given by 𝑢(𝑥)𝑣(𝑥)=𝑢(𝑥)𝑣(𝑥)𝑢(𝑥)𝑣(𝑥)(𝑣(𝑥)).

For 𝑦, we can switch out the 𝑥 variable for 𝑧, and we have 𝑢(𝑧)=𝑧+3 and 𝑣(𝑧)=𝑧+13. The derivative of these functions with the power rule is straightforward; we get 𝑢(𝑧)=1 and 𝑣(𝑧)=1. Combining this all into the quotient rule equation, we get dd𝑦𝑧=𝑢(𝑧)𝑣(𝑧)=1(𝑧+13)(𝑧+3)1(𝑧+13)=10(𝑧+13).

We can perform a very similar calculation for dd𝑧𝑥. Let 𝑧=𝑓(𝑥)𝑔(𝑥), so that 𝑓(𝑥)=𝑥10 and 𝑔(𝑥)=𝑥3, with derivatives 𝑓(𝑥)=1 and 𝑔(𝑥)=1. Then, using the quotient rule, we get dd𝑧𝑥=1(𝑥3)(𝑥10)1(𝑥3)=7(𝑥3).

Returning to our formula for the chain rule, we multiply these two derivatives together to get dd𝑦𝑥=10(𝑧+13)7(𝑥3)=70(𝑧+13)(𝑥3).

We could express this purely in terms of 𝑥, but this is unnecessary since we only need to evaluate the derivative at the given point, 𝑥=4. Note that at this point, 𝑧=41043=6. Thus, letting 𝑥=4 and 𝑧=6, we have dd𝑦𝑥|||=70(6+13)(43)=107.

Let us recap a few important points that we have learned in this explainer.

Key Points

  • Using the rules of differentiation, namely, the product, quotient, and chain rules, we can calculate the derivatives of any combination of elementary functions.
  • It is important to consider the order in which we use the rules as this will help ensure we choose the most efficient method.
  • Generally, we want to consider the outermost parts of the function and work inward. Doing this, we can decompose a complicated function into simpler parts that can be evaluated directly.
  • It is worth keeping in mind that some problems can be simplified to the point where we do not need to use multiple rules of differentiation, making the process significantly easier (even though such problems have not been featured in this explainer).

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