The portal has been deactivated. Please contact your portal admin.

Lesson Explainer: Reverse Chain Rule Mathematics

In this explainer, we will learn how to evaluate integrals of functions in the form 𝑘𝑓′(𝑥)𝑓(𝑥) and 𝑘𝑓′(𝑥)(𝑓(𝑥)).

Integration and differentiation are the reverse processes of each other. This means that knowing how to perform differentiation on certain expressions can help solve certain types of integration problems. Furthermore, we can verify our answer to an integration problem by differentiating it. In this sense, a very useful application of this property is that any derivative result can be stated as an integration result by reversing the process.

For example, we recall that the chain rule tells us that if 𝑓 is differentiable at 𝑥 and 𝑔 is differentiable at 𝑓(𝑥), then dd𝑥(𝑔(𝑓(𝑥)))=𝑓′(𝑥)𝑔′(𝑓(𝑥)).

We can reverse this process by integrating both sides of this result with respect to 𝑥. We note that integrating a derivative leaves the function unchanged: 𝑥(𝑔(𝑓(𝑥)))𝑥=𝑓′(𝑥)𝑔′(𝑓(𝑥))𝑥𝑔(𝑓(𝑥))=𝑓′(𝑥)𝑔′(𝑓(𝑥))𝑥.ddddd

This is now in the form of an integral result, where we need to add a constant of integration as usual: 𝑓′(𝑥)𝑔′(𝑓(𝑥))𝑥=𝑔(𝑓(𝑥))+𝐶.d

This is known as the reverse chain rule since it is found by reversing the chain rule by integration.

Theorem: Reverse Chain Rule

If 𝑓 is differentiable at 𝑥 and 𝑔 is differentiable at 𝑓(𝑥), then 𝑓′(𝑥)𝑔′(𝑓(𝑥))𝑥=𝑔(𝑓(𝑥))+𝐶.d

There are many applications of the chain rule; however, in this explainer, we will focus on two specific applications of this result.

Let 𝑔(𝑥)=𝑥 for some unknown constant 𝑛 and let 𝑓(𝑥) be a differentiable function. We can recall that the power rule for differentiation tells us that 𝑔′(𝑥)=(𝑛+1)𝑥. Substituting these expressions into the reverse chain rule, we obtain 𝑓′(𝑥)(𝑛+1)(𝑓(𝑥))𝑥=(𝑓(𝑥))+𝐶.d

We want to take out the factor of 𝑛+1 from the integral and divide both sides of the equation by 𝑛+1; however, we can only do this if 𝑛≠−1: 𝑓′(𝑥)(𝑓(𝑥))𝑥=1𝑛+1(𝑓(𝑥))+𝐶.d

Multiplying the constant of integration by a constant yields a constant, so we have 𝑓′(𝑥)(𝑓(𝑥))𝑥=1𝑛+1(𝑓(𝑥))+𝐶.d

We can generalize this by multiplying both sides of the equation by a constant 𝑘: 𝑘𝑓′(𝑥)(𝑓(𝑥))𝑥=𝑘𝑛+1(𝑓(𝑥))+𝐶.d

We have the following result.

Property: Application of the Reverse Chain Rule

For any differentiable function 𝑓(𝑥) and real constants 𝑘 and 𝑛, where 𝑛≠−1, we have 𝑘𝑓′(𝑥)(𝑓(𝑥))𝑥=𝑘𝑛+1(𝑓(𝑥))+𝐶.d

This allows us to evaluate any integral in this form. For example, consider 4𝑥𝑥−7𝑥.d

We could evaluate this integral by expanding the brackets using the binomial expansion formula; however, it is easier to set 𝑓(𝑥)=𝑥−7 in the reverse chain rule formula. We then have 𝑓′(𝑥)=2𝑥, and we can note that 4𝑥=2(2𝑥)=2𝑓′(𝑥).

Therefore, if we substitute 𝑘=2, 𝑛=8, and 𝑓(𝑥)=𝑥−7 into the reverse chain rule formula, we obtain 4𝑥𝑥−7𝑥=2𝑓′(𝑥)(𝑓(𝑥))𝑥=28+1𝑥−7+𝐶=29𝑥−7+𝐶.dd

In our first example, we will apply the reverse chain rule to integrate a rational function.

Example 1: Using the Reverse Chain Rule to Integrate a Function

Determine 6𝑥+83𝑥+8𝑥+3𝑥d.

Answer

In order to answer this question, we first note that we are asked to integrate the quotient of two polynomials. We can start by checking for patterns in the integrand first to see if this is in a standard form for integration. One thing we can always check is whether we can apply the reverse chain rule, which states that for any constant 𝑘 and differentiable function 𝑓(𝑥), 𝑘𝑓′(𝑥)𝑓(𝑥)𝑥=𝑘|𝑓(𝑥)|+𝐶.dln

To check if we can apply this result, we need to differentiate the denominator in the integrand. We call this function 𝑓(𝑥) and differentiate it term by term using the power rule for differentiation: 𝑓′(𝑥)=3𝑥+8𝑥+3=6𝑥+8.

We see that this is exactly equal to the numerator of the integrand, so we have 6𝑥+83𝑥+8𝑥+3𝑥=𝑓′(𝑥)𝑓(𝑥)𝑥=|𝑓(𝑥)|+𝐶=||3𝑥+8𝑥+3||+𝐶.ddlnln

Let’s verify our answer by differentiation. We recall that for a differentiable function 𝑓(𝑥), ddln𝑥(|𝑓(𝑥)|)=𝑓′(𝑥)𝑓(𝑥).

Therefore, ddlnddlndd𝑥||3𝑥+8𝑥+3||+𝐶=𝑥||3𝑥+8𝑥+3||+𝑥(𝐶)=6𝑥+83𝑥+8𝑥+3.

This is equal to the integrand we are given, so it is the correct antiderivative. This verifies that the answer is correct.

Hence, 6𝑥+83𝑥+8𝑥+3𝑥=||3𝑥+8𝑥+3||+𝐶.dln

In our next example, we will apply the reverse chain rule to integrate the quotient of two functions.

Example 2: Using the Reverse Chain Rule to Integrate a Function

Determine −9𝑒7𝑒+12𝑥d.

Answer

In order to answer this question, we first note that we are asked to integrate the quotient of two functions. Integrating the quotient of functions is difficult, so we should always check for patterns in the integrand first to look for the simplest method of integration.

Let’s check if we can apply the reverse chain rule, which states that for any constant 𝑘 and differentiable function 𝑓(𝑥), 𝑘𝑓′(𝑥)𝑓(𝑥)𝑥=𝑘|𝑓(𝑥)|+𝐶.dln

To apply this result, we need to differentiate the function in the denominator of the integrand and compare it to the function in the numerator of the integrand. We can do this by calling the function in the denominator of the integrand 𝑓(𝑥) and recalling that dd𝑥(𝑒)=𝑒. Therefore, 𝑓′(𝑥)=𝑥(7𝑒+12)=7𝑒.dd

This is not equal to the numerator of the integrand. However, it is a constant multiple of the numerator of the integrand: 97(7𝑒)=9𝑒.

We can use this to apply the reverse chain rule to evaluate the integral. We first take the factor of −1 out of the integral: −9𝑒7𝑒+12𝑥=−9𝑒7𝑒+12𝑥.dd

Next, we can set 𝑓(𝑥)=7𝑒+12; then, 𝑓′(𝑥)=7𝑒. We rewrite 9𝑒 as 97(7𝑒)=97𝑓′(𝑥) to get −9𝑒7𝑒+12𝑥=−(7𝑒)𝑓(𝑥)𝑥=−𝑓′(𝑥)𝑓(𝑥)𝑥.ddd

This is now in the form of the reverse chain rule with 𝑘=97, so we have −𝑓′(𝑥)𝑓(𝑥)𝑥=−97|𝑓(𝑥)|+𝐶.dln

We can substitute 𝑓(𝑥)=7𝑒+12 into the result, and we note that 𝐶 is just a constant, so −𝐶 is also a constant. We can just call the final constant 𝐶: −97|𝑓(𝑥)|+𝐶=−97|7𝑒+12|+𝐶.lnln

A second application of the reverse chain rule is to let 𝑔(𝑥)=|𝑥|ln, so we have 𝑔′(𝑥)=1𝑥. From the reverse chain rule, we have 𝑓′(𝑥)𝑔′(𝑓(𝑥))𝑥=𝑔(𝑓(𝑥))+𝐶𝑓′(𝑥)1𝑓(𝑥)𝑥=|𝑓(𝑥)|+𝐶.ddln

Rearranging and multiplying through by a constant 𝑘 yields 𝑘𝑓′(𝑥)𝑓(𝑥)𝑥=𝑘|𝑓(𝑥)|+𝐶.dln

We have shown the following two applications of the reverse chain rule.

Property: Application of the Reverse Chain Rule

For any differentiable function 𝑓(𝑥) and real constant 𝑘, we have 𝑘𝑓′(𝑥)𝑓(𝑥)𝑥=𝑘|𝑓(𝑥)|+𝐶.dln

In our next example, we will apply the reverse chain rule to integrate the product of trigonometric functions.

Example 3: Using the Reverse Chain Rule to Integrate a Function

Use the reverse chain rule to find an expression for −(𝑥)(𝑥)𝑥cossind.

Answer

Let’s begin by recalling the version of the reverse chain rule that is most relevant to our integral: for a differentiable function 𝑓(𝑥) and constants 𝑘 and 𝑛, with 𝑛≠−1, 𝑘𝑓′(𝑥)(𝑓(𝑥))𝑥=𝑘𝑛+1(𝑓(𝑥))+𝐶.d

We can set 𝑓(𝑥)=𝑥cos, since this is the function raised to a power. We then have 𝑓′(𝑥)=−𝑥sin. We can then rewrite our integral in this form: −(𝑥)(𝑥)𝑥=−(𝑥)(𝑥)𝑥=𝑓′(𝑥)(𝑓(𝑥))𝑥.cossindsincosdd

Therefore, we set 𝑘=1, 𝑛=4, and 𝑓(𝑥)=𝑥cos and apply the reverse chain rule to get (𝑥)(𝑥)𝑥=14+1((𝑥))+𝐶=15(𝑥)+𝐶.cossindcoscos

In our next example, we will use the reverse chain rule to evaluate a definite integral.

Example 4: Using the Reverse Chain Rule to Evaluate an Integral of a Function

Use the reverse chain rule to evaluate 18𝑥−12√𝑥−2𝑥+1𝑥d.

Answer

Let’s begin by recalling the version of the reverse chain rule that is most relevant to our integral: for a differentiable function 𝑓(𝑥) and constants 𝑘 and 𝑛, with 𝑛≠−1, 𝑘𝑓′(𝑥)(𝑓(𝑥))𝑥=𝑘𝑛+1(𝑓(𝑥))+𝐶.d

We know that taking the square root is the same as raising the expression to a power of 12, so we have 18𝑥−12√𝑥−2𝑥+1𝑥=18𝑥−12𝑥−2𝑥+1𝑥.dd

To apply the reverse chain rule, we need to set 𝑓(𝑥)=𝑥−2𝑥+1, and since this is the term raised to a power, we can differentiate 𝑓(𝑥) term by term by using the power rule for differentiation to get 𝑓′(𝑥)=3𝑥−2.

We want to compare this to 18𝑥−12; we can see that 18𝑥−12=63𝑥−2=6𝑓′(𝑥).

Therefore, we can rewrite the integral as 18𝑥−12√𝑥−2𝑥+1𝑥=6𝑓′(𝑥)(𝑓(𝑥))𝑥.dd

This is in the form of the reverse chain rule with 𝑘=6, 𝑛=12, and 𝑓(𝑥)=𝑥−2𝑥+1. Hence, by applying the reverse chain rule, we have 18𝑥−12√𝑥−2𝑥+1𝑥=6+1𝑥−2𝑥+1.d

We calculate that 6+1=4 and take this factor out of the evaluation and simplify to get 6+1𝑥−2𝑥+1=4𝑥−2𝑥+1=4𝑥−2𝑥+1.

We now need to evaluate at the limits of integration: 4𝑥−2𝑥+1=40−2(0)+1−(−1)−2(−1)+1=41−2=41−2√2=4−8√2.

Hence, 18𝑥−12√𝑥−2𝑥+1𝑥=4−8√2.d

In our final example, we will use the reverse chain rule to evaluate the definite integral of a function in order to determine the upper bound of the integral.

Example 5: Using the Reverse Chain Rule to Find the Value of a Constant in a Definite Integral

Given that 3𝑥𝑥𝑥=2√2−1sectand and 0<𝜃<𝜋2, find the value of 𝜃.

Answer

To find the value of 𝜃, we first need to find an expression for 3𝑥𝑥𝑥sectand in terms of 𝜃. We can do this by noting that the integrand is the product of two functions, so we can try applying the reverse chain rule, which states that for a differentiable function 𝑓(𝑥) and constants 𝑘 and 𝑛, with 𝑛≠−1, 𝑘𝑓′(𝑥)(𝑓(𝑥))𝑥=𝑘𝑛+1(𝑓(𝑥))+𝐶.d

We can set 𝑓(𝑥)=𝑥sec, since this is the function that is raised to a power. We then recall that 𝑓′(𝑥)=𝑥𝑥tansec, which appears in the integral. This allows us to rewrite the integral as 3𝑥𝑥𝑥=3(𝑥𝑥)𝑥𝑥.sectandtansecsecd

This is now in the form of the reverse chain rule with 𝑘=3, 𝑛=2, and 𝑓(𝑥)=𝑥sec. Therefore, 3𝑥𝑥𝑥=32+1(𝑥)=𝑥.sectandsecsec

We can now evaluate this at the limits of integration to obtain 𝑥=𝜃−0=𝜃−1.secsecsecsec

Hence, 3𝑥𝑥𝑥=𝜃−1.sectandsec

We are given that 3𝑥𝑥𝑥=2√2−1sectand, so our expression for this integral must be equal to this value: sec𝜃−1=2√2−1.

Adding 1 to both sides yields sec𝜃=2√2.

Taking the cube root of both sides of the equation gives us sec𝜃=√2.

Taking the reciprocal of both sides of the equation and recalling that 1𝜃=𝜃seccos gives cos𝜃=1√2=√22.

We are given that 0<𝜃<𝜋2, and there is only one solution in this interval for the equation; that is, 𝜃=√22=𝜋4.cos

Hence, 𝜃=𝜋4.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The reverse chain rule states that for differentiable functions 𝑓(𝑥) and 𝑔(𝑥), 𝑓′(𝑥)𝑔′(𝑓(𝑥))𝑥=𝑔(𝑓(𝑥))+𝐶.d
  • Applying the reverse chain rule with 𝑔(𝑥)=𝑥 and 𝑛≠−1 yields 𝑘𝑓′(𝑥)(𝑓(𝑥))𝑥=𝑘𝑛+1(𝑓(𝑥))+𝐶.d
  • Applying the reverse chain rule with 𝑔(𝑥)=|𝑥|ln yields 𝑘𝑓′(𝑥)𝑓(𝑥)𝑥=𝑘|𝑓(𝑥)|+𝐶.dln

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.