Explainer: Differentiation of Inverse Functions

In this explainer, we will learn how to find the derivatives of inverse functions.

We will begin by recapping some important facts about inverse functions.

Inverse Function

Let 𝑓 be a function with domain π‘ˆ and range 𝑉. We call a function π‘”βˆΆπ‘‰β†’π‘ˆ the inverse of 𝑓 if, for all π‘¦βˆˆπ‘‰, 𝑓(𝑔(𝑦))=𝑦, and, for all π‘₯βˆˆπ‘ˆ, 𝑔(𝑓(π‘₯))=π‘₯.

When we can find such a function 𝑔, we say 𝑓 is invertible and 𝑔 is the inverse of 𝑓. The inverse of a function is unique. We often use the notation π‘“οŠ±οŠ§ to represent the inverse of a function. Additionally, if 𝑔 is the inverse of 𝑓, then 𝑓 is the inverse of 𝑔.

Recall that if we plot the graph of a function, we can find the graph of its inverse by reflecting the graph in the line 𝑦=π‘₯.

If we consider a tangent to the graph of 𝑓(π‘₯), its reflection in the line 𝑦=π‘₯ will be tangent to the graph of 𝑓(π‘₯).

This tells us that there should be some algebraic relationship between the slope of the tangent to the graph of 𝑓 and the slope of the tangent to the graph of 𝑓(π‘₯). In fact, we can be even more explicit about what this relationship should be. Let 𝐿 be the tangent to 𝑓(π‘₯) and 𝐿 be its image under reflection in the line 𝑦=π‘₯. We can write the equation of 𝐿 as follows: 𝑦=π‘šπ‘₯+𝑐, where π‘š is the slope of the line. We now reflect 𝐿 in the line 𝑦=π‘₯. This corresponds to mapping (π‘₯,𝑦) to (𝑦,π‘₯).Hence, the equation of the line 𝐿 can be written as π‘₯=π‘šπ‘¦+𝑐.

Rearranging this, we can put this into slope–intercept form. Hence, by subtracting 𝑐 from both sides, we have π‘šπ‘¦=π‘₯βˆ’π‘.

Now we can divide by π‘š to get 𝑦=1π‘šπ‘₯βˆ’π‘π‘š.

This is an equation in slope–intercept form. Hence, we can immediately read off its slope to be 1π‘š. What we have seen is that the slopes of the tangents to the graphs of 𝑓 and π‘“οŠ±οŠ§ are reciprocals of one another.

Since the derivative represents the slope of the tangent at each point on the curve, we would expect that if we know the derivative of 𝑓 at some point π‘Ž, then the derivative of π‘“οŠ±οŠ§ at 𝑏=𝑓(π‘Ž) will be its reciprocal. Hence, we anticipate that 𝑓(𝑏)=1𝑓(π‘Ž), which only makes sense if 𝑓(π‘Ž)β‰ 0. This statement is of course true and is easy to prove using the chain rule, as we shall demonstrate.

Consider the function 𝑓 with an inverse 𝑔; then, by definition, 𝑓(𝑔(𝑦))=𝑦.

Using the chain rule, we can differentiate both sides with respect to 𝑦 as follows: 𝑓(𝑔(𝑦))𝑔(𝑦)=1.

Hence, if 𝑓(𝑔(𝑦))β‰ 0, we have 𝑔(𝑦)=1𝑓(𝑔(𝑦)).

This is often written using Leibniz’s notation as dd𝑦π‘₯=1.ddο—ο˜

We will now consider some examples where we use this result to find the derivatives of a function and their inverses. In the first example, we will consider a simple function that we know how to differentiate and to find its inverse.

Example 1: Verifying the Formula for the Derivative of an Inverse Function

Consider the function 𝑦=π‘₯, where π‘₯β‰₯0.

  1. Find dd𝑦π‘₯.
  2. Express 1ddο˜ο— in terms of 𝑦.
  3. By expressing π‘₯ in terms of 𝑦, find an expression for ddπ‘₯𝑦.

Answer

Part 1

Using the power rule, we have dd𝑦π‘₯=2π‘₯.

Part 2

By taking reciprocals, we have 1=12π‘₯.ddο˜ο—

Using the equation of the function 𝑦=π‘₯, we can express π‘₯ in terms of 𝑦 as π‘₯=βˆšπ‘¦. Substituting this into the equation, we have 1=12βˆšπ‘¦.ddο˜ο—

Part 3

Since π‘₯=βˆšπ‘¦, we can now differentiate this with respect to 𝑦 to get ddπ‘₯𝑦=12βˆšπ‘¦.

The previous example confirmed that ddπ‘₯𝑦=1,ddο˜ο— for the given function. However, it is interesting to note that the derivative ddπ‘₯𝑦 is not defined when π‘₯=0. This is as a direct result of the fact that dd𝑦π‘₯=0 at this point.

Using the inverse function theorem, we can find the derivative of familiar functions such as the natural logarithm or inverse trigonometric functions.

Example 2: Finding the Derivatives of Familiar Inverse Functions

Given that π‘₯=π‘’ο˜, find dd𝑦π‘₯, giving your answer in terms of π‘₯.

Answer

We begin by taking the derivative of π‘₯ with respect to 𝑦. Using the rule for differentiating the exponential function, we have ddπ‘₯𝑦=𝑒.

Since 𝑒>0 for all 𝑦, we can find the derivative of the inverse function dd𝑦π‘₯ by taking the reciprocal as follows: dd𝑦π‘₯=1=1𝑒.ddο—ο˜ο˜

Since π‘₯=π‘’ο˜, we can rewrite this as dd𝑦π‘₯=1π‘₯.

The last example showed us that ddlnπ‘₯π‘₯=1π‘₯.

We can use a similar technique to find formulae for the derivatives of inverse sine, cosine, and tangent.

Sometimes, we will be asked to find the derivative of the inverse of a function at a given point. We can use a similar technique; however, we need to be careful about the point we evaluate the function as the next example will demonstrate.

Example 3: Finding the Derivatives of Inverse Functions

Let 𝑓(π‘₯)=12π‘₯+12π‘₯+5π‘₯βˆ’4 and let 𝑔 be the inverse of 𝑓. Given that 𝑓(2)=12, what is 𝑔(12)?

Answer

We start by finding the derivative of 𝑓 using the power rule to get 𝑓(π‘₯)=32π‘₯+π‘₯+5.

We can now use the formula for the derivative of the inverse 𝑔, 𝑔(𝑦)=1𝑓(𝑔(𝑦)), to find the derivative at π‘₯=12. To find the derivative at this point, we need to know the value of 𝑔(12). Since 𝑓(2)=12, we can apply 𝑔 to both sides of the equation to get 𝑔(𝑓(2))=𝑔(12). Since 𝑔 is the inverse of 𝑓, we can rewrite this as 2=𝑔(12). Hence, 𝑔(12)=1𝑓(2).

Therefore, using the expression for the derivative of 𝑓, we have 𝑔(12)=1(2)+(2)+5=113.

Clearly, in the last couple of examples, we were able to find explicit representation of the inverse of the function. However, this is not always the case, as we shall see in the next couple of examples.

Example 4: Finding the Derivatives of Inverse Functions

Given that π‘₯=𝑒+π‘¦ο˜sin, find dd𝑦π‘₯.

Answer

In an example like this, we are clearly unable to find an explicit representation for 𝑦. However, we can use the following formula for the derivative of the inverse function: dd𝑦π‘₯=1.ddο—ο˜

Therefore, we begin by calculating the derivative ddπ‘₯𝑦 as follows: ddcosπ‘₯𝑦=𝑒+𝑦.

Hence, ddcos𝑦π‘₯=1𝑒+𝑦.

Example 5: Finding the Derivatives of Inverse Functions

Given that π‘₯=𝑦+βˆšπ‘¦+βˆšπ‘¦οŠ«οŠ¨οŽ’, find dd𝑦π‘₯.

Answer

Using the formula for the derivative of the inverse function, dd𝑦π‘₯=1,ddο—ο˜ we can find the derivative dd𝑦π‘₯. We start by differentiating the expression for π‘₯ with respect to 𝑦 to get ddπ‘₯𝑦=5𝑦+12βˆšπ‘¦+23βˆšπ‘¦=30𝑦+4βˆšπ‘¦+3βˆšπ‘¦6𝑦.οŠͺ

Hence, dd𝑦π‘₯=6𝑦30𝑦+4βˆšπ‘¦+3βˆšπ‘¦.

Moreover, we can even find the derivatives of inverse functions without an algebraic expression defining the function. In the next example, we will see how we can use a table of function values to find the derivative of the inverse function.

Example 6: Derivatives of Inverse Functions Using Tables

Let 𝑔 be the inverse of 𝑓.

Using the values in the table below, find 𝑔(1).

π‘₯𝑓(π‘₯)𝑔(π‘₯)𝑓(π‘₯)
1βˆ’52βˆ’2
2βˆ’941

Answer

Given that dd𝑦π‘₯=1,ddο—ο˜ we might be tempted to say that 𝑔(1)=1𝑓(1); however, this is incorrect. Therefore, we need to be careful to evaluate π‘“οŽ˜ at the correct point, which is not where π‘₯=1, but where π‘₯=𝑔(1). The best way to remember this is using the formula 𝑔(𝑦)=1𝑓(𝑔(𝑦)).

Substituting in 𝑦=1, we have 𝑔(1)=1𝑓(𝑔(1)).

Referring to the table, we see that 𝑔(1)=2. Therefore, 𝑔(1)=1𝑓(2).

Hence, using the value from the table, we have 𝑔(1)=11=1.

Example 7: Derivatives of Inverse Functions Using Tables

Let 𝑔 be the inverse of 𝑓.

Using the values in the table below, find 𝑔(0).

π‘₯𝑓(π‘₯)𝑔(π‘₯)𝑓(π‘₯)
βˆ’1 5 913
0 3 βˆ’10

Answer

Recall that the derivative of the inverse of a function can be found using the following formula: 𝑔(𝑦)=1𝑓(𝑔(𝑦)).

We would like to know the derivative at the point where 𝑦=0; therefore, 𝑔(0)=1𝑓(𝑔(0)).

Using the values from the table, we can see that 𝑔(0)=βˆ’1. Hence, 𝑔(0)=1𝑓(βˆ’1)=1=3.

The statement which we proved earlier, that the derivative of the inverse of a function is the reciprocal of the derivative of the function, assumed that the inverse existed. However, the full statement of the inverse function theorem is actually much more powerful in that it guarantees the existence and continuity of the inverse of a function when it is continuously differentiable with a nonzero derivative. The full statement of the theorem is below.

Inverse Function Theorem

Let 𝑓 be a continuously differentiable function with a nonzero derivative at a point π‘Ž. Then, the inverse function theorem tells us that

  1. 𝑓 is invertible in a neighborhood of π‘Ž;
  2. 𝑓 has a continuously differentiable inverse in a neighborhood of π‘Ž;
  3. the derivative of the inverse at the point 𝑏=𝑓(π‘Ž) is equal to the reciprocal of the derivative of 𝑓 at π‘Ž; that is, 𝑓(𝑏)=1𝑓(π‘Ž).

The proof of this theorem is rather technical and requires advanced knowledge of analysis appealing to the completeness of the real numbers, the intermediate value theorem, and so on. It is, therefore, beyond the scope of this explainer. However, it is extremely useful since it guarantees the existence of the inverse at particular points.

Key Points

  • Given a continuously differentiable function 𝑓 with a nonzero derivative at a point π‘Ž, the derivative of the inverse function at 𝑏=𝑓(π‘Ž) is 𝑓(𝑏)=1𝑓(π‘Ž). This is often written in Leibniz’s notation as dd𝑦π‘₯=1.ddο—ο˜
  • We need to be careful about which points we are using.
  • When using these equations, we can find the derivatives of many familiar inverse functions, such as the natural logarithm and inverse trigonometric function.
  • The inverse function theorem guarantees the existence of the inverse of a continuous function around points with nonzero derivatives.
  • Using this theorem, we can find the derivative of inverse functions even when we are unable to find an explicit formula for the inverse.

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