Ulearngo

The Derivative of an Inverse Function

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

\[\left(\right. f^{−1} \left.\right)^{'} \left(\right. a \left.\right) = \frac{1}{f^{'} \left(\right. f^{−1} \left(\right. a \left.\right) \left.\right)} .\]
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 3.28 The 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 \(x = f \left(\right. f^{−1} \left(\right. x \left.\right) \left.\right) .\) Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

\[1 = f^{'} \left(\right. f^{−1} \left(\right. x \left.\right) \left.\right) \left(\right. f^{−1} \left.\right)^{'} \left(\right. x \left.\right) .\]

Solving for \(\left(\right. f^{−1} \left.\right)^{'} \left(\right. x \left.\right) ,\) we obtain

\[\left(\right. f^{−1} \left.\right)^{'} \left(\right. x \left.\right) = \frac{1}{f^{'} \left(\right. f^{−1} \left(\right. x \left.\right) \left.\right)} .\]

We summarize this result in the following theorem.

Theorem 3.11

Inverse Function Theorem

Let \(f \left(\right. x \left.\right)\) be a function that is both invertible and differentiable. Let \(y = f^{−1} \left(\right. x \left.\right)\) be the inverse of \(f \left(\right. x \left.\right) .\) For all \(x\) satisfying \(f^{'} \left(\right. f^{−1} \left(\right. x \left.\right) \left.\right) \neq 0 ,\)

\[\frac{d y}{d x} = \frac{d}{d x} \left(\right. f^{−1} \left(\right. x \left.\right) \left.\right) = \left(\right. f^{−1} \left.\right)^{'} \left(\right. x \left.\right) = \frac{1}{f^{'} \left(\right. f^{−1} \left(\right. x \left.\right) \left.\right)} .\]

Alternatively, if \(y = g \left(\right. x \left.\right)\) is the inverse of \(f \left(\right. x \left.\right) ,\) then

\[g ' \left(\right. x \left.\right) = \frac{1}{f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right)} .\]

Example 3.60

Applying the Inverse Function Theorem

Use the inverse function theorem to find the derivative of \(g \left(\right. x \left.\right) = \frac{x + 2}{x} .\) Compare the resulting derivative to that obtained by differentiating the function directly.

Solution

The inverse of \(g \left(\right. x \left.\right) = \frac{x + 2}{x}\) is \(f \left(\right. x \left.\right) = \frac{2}{x - 1} .\) Since \(g^{'} \left(\right. x \left.\right) = \frac{1}{f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right)} ,\) begin by finding \(f^{'} \left(\right. x \left.\right) .\) Thus,

\[f^{'} \left(\right. x \left.\right) = \frac{−2}{\left(\right. x - 1 \left.\right)^{2}} \text{and} f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right) = \frac{−2}{\left(\right. g \left(\right. x \left.\right) - 1 \left.\right)^{2}} = \frac{−2}{\left(\right. \frac{x + 2}{x} - 1 \left.\right)^{2}} = - \frac{x^{2}}{2} .\]

Finally,

\[g^{'} \left(\right. x \left.\right) = \frac{1}{f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right)} = - \frac{2}{x^{2}} .\]

We can verify that this is the correct derivative by applying the quotient rule to \(g \left(\right. x \left.\right)\) to obtain

\[g^{'} \left(\right. x \left.\right) = - \frac{2}{x^{2}} .\]

Checkpoint 3.42

Use the inverse function theorem to find the derivative of \(g \left(\right. x \left.\right) = \frac{1}{x + 2} .\) Compare the result obtained by differentiating \(g \left(\right. x \left.\right)\) directly.

Example 3.61

Applying the Inverse Function Theorem

Use the inverse function theorem to find the derivative of \(g \left(\right. x \left.\right) = \sqrt[3]{x} .\)

Solution

The function \(g \left(\right. x \left.\right) = \sqrt[3]{x}\) is the inverse of the function \(f \left(\right. x \left.\right) = x^{3} .\) Since \(g^{'} \left(\right. x \left.\right) = \frac{1}{f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right)} ,\) begin by finding \(f^{'} \left(\right. x \left.\right) .\) Thus,

\[f^{'} \left(\right. x \left.\right) = 3 x^{2} \text{and} f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right) = 3 \left(\right. \sqrt[3]{x} \left.\right)^{2} = 3 x^{2 / 3} .\]

Finally,

\[g^{'} \left(\right. x \left.\right) = \frac{1}{3 x^{2 / 3}} = \frac{1}{3} x^{−2 / 3} .\]

Checkpoint 3.43

Find the derivative of \(g \left(\right. x \left.\right) = \sqrt[5]{x}\) by applying the inverse function 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.

Theorem 3.12

Extending the Power Rule to Rational Exponents

The power rule may be extended to rational exponents. That is, if \(n\) is a positive integer, then

\[\frac{d}{d x} \left(\right. x^{1 / n} \left.\right) = \frac{1}{n} x^{\left(\right. 1 / n \left.\right) - 1} .\]

Also, if \(n\) is a positive integer and \(m\) is an arbitrary integer, then

\[\frac{d}{d x} \left(\right. x^{m / n} \left.\right) = \frac{m}{n} x^{\left(\right. m / n \left.\right) - 1} .\]

Proof

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

\[f^{'} \left(\right. x \left.\right) = n x^{n - 1} \text{and} f^{'} \left(\right. g \left(\right. x \left.\right) \left.\right) = n \left(\left(\right. x^{1 / n} \left.\right)\right)^{n - 1} = n x^{\left(\right. n - 1 \left.\right) / n} .\]

Finally,

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

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

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

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Example 3.62

Applying the Power Rule to a Rational Power

Find the equation of the line tangent to the graph of \(y = x^{2 / 3}\) at \(x = 8 .\)

Solution

First find \(\frac{d y}{d x}\) and evaluate it at \(x = 8 .\) Since

\[\begin{aligned} \frac{d y}{d x} = \frac{2}{3} x^{−1 / 3} \text{and} \frac{d y}{d x} \left|\right. \\ _{x = 8} = \frac{1}{3} \end{aligned}\]

the slope of the tangent line to the graph at \(x = 8\) is \(\frac{1}{3} .\)

Substituting \(x = 8\) into the original function, we obtain \(y = 4 .\) Thus, the tangent line passes through the point \(\left(\right. 8 , 4 \left.\right) .\) Substituting into the point-slope formula for a line, we obtain the tangent line

\[y = \frac{1}{3} x + \frac{4}{3} .\]

Checkpoint 3.44

Find the derivative of \(s \left(\right. t \left.\right) = \sqrt{2 t + 1} .\)

This lesson is part of:

Derivatives

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