3.7 Derivatives of Inverse Functions

In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functions whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric functions. This formula may also be used to extend the power rule to rational exponents.

The Derivative of an Inverse Function

We begin by considering a function and its inverse. If $f (x)$ is both invertible and differentiable, it seems reasonable that the inverse of $f (x)$ is also differentiable. Figure 3.28 shows the relationship between a function $f (x)$ and its inverse $f^{- 1} (x) .$ Look at the point $(a, f^{- 1} (a))$ on the graph of $f^{- 1} (x)$ having a tangent line with a slope of $(f^{- 1})^{'} (a) = \frac{p}{q} .$ This point corresponds to a point $(f^{- 1} (a), a)$ on the graph of $f (x)$ having a tangent line with a slope of $f^{'} (f^{- 1} (a)) = \frac{q}{p} .$ Thus, if $f^{- 1} (x)$ is differentiable at $a,$ then it must be the case that

(f^{- 1})^{'} (a) = \frac{1}{f ^{'} ( f ^{- 1} ( a ))} .

This graph shows a function f(x) and its inverse f−1(x). These functions are symmetric about the line y = x. The tangent line of the function f(x) at the point (f−1(a), a) and the tangent line of the function f−1(x) at (a, f−1(a)) are also symmetric about the line y = x. Specifically, if the slope of one were p/q, then the slope of the other would be q/p. Lastly, their derivatives are also symmetric about the line y = x. — Figure 3.28The tangent lines of a function and its inverse are related; so, too, are the derivatives of these functions.

We may also derive the formula for the derivative of the inverse by first recalling that $x = f (f^{- 1} (x)) .$ Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

1 = f^{'} (f^{- 1} (x)) (f^{- 1})^{'} (x) .

Solving for $(f^{- 1})^{'} (x),$ we obtain

(f^{- 1})^{'} (x) = \frac{1}{f ^{'} ( f ^{- 1} ( x ))} .

We summarize this result in the following theorem.

From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form $\frac{1}{n},$ where $n$ is a positive integer. This extension will ultimately allow us to differentiate $x^{q},$ where $q$ is any rational number.

Proof

The function $g (x) = x^{1 / n}$ is the inverse of the function $f (x) = x^{n} .$ Since $g^{'} (x) = \frac{1}{f ^{'} ( g ( x ))},$ begin by finding Thus,

f^{'} (x) = n x^{n - 1} and f^{'} (g (x)) = n (x^{1 / n})^{n - 1} = n x^{(n - 1) / n} .

Finally,

g^{'} (x) = \frac{1}{n x ^{(n - 1) / n}} = \frac{1}{n} x^{(1 - n) / n} = \frac{1}{n} x^{(1 / n) - 1} .

To differentiate $x^{m / n}$ we must rewrite it as $(x^{1 / n})^{m}$ and apply the chain rule. Thus,

\frac{d}{d x} (x^{m / n}) = \frac{d}{d x} ((x^{1 / n})^{m}) = m (x^{1 / n})^{m - 1} \cdot \frac{1}{n} x^{(1 / n) - 1} = \frac{m}{n} x^{(m / n) - 1} .

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Derivatives of Inverse Trigonometric Functions

We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.

The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.

Key Concepts

The inverse function theorem allows us to compute derivatives of inverse functions without using the limit definition of the derivative.
We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.

Key Equations

Inverse function theorem

(f^{- 1})^{'} (x) = \frac{1}{f ^{'} ( f ^{- 1} ( x ))}

whenever and is differentiable.