Derivatives of Various Inverse Functions

Learning Outcomes

Calculate 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 1 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 $(f^{-1})^{\prime}(a)=\frac{p}{q}$ . This point corresponds to a point $(f^{-1}(a),a)$ on the graph of $f(x)$ having a tangent line with a slope of $f^{\prime}(f^{-1}(a))=\frac{q}{p}$ . 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))}

$(f^{-1})^{\prime}(a)=\dfrac{1}{f^{\prime}(f^{-1}(a))}$

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

$1=f^{\prime}(f^{-1}(x))(f^{-1})^{\prime}(x))$

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

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

$(f^{-1})^{\prime}(x)=\dfrac{1}{f^{\prime}(f^{-1}(x))}$

We summarize this result in the following theorem.

Inverse Function Theorem

Let $f(x)$ be a function that is both invertible and differentiable. Let $y=f^{-1}(x)$ be the inverse of $f(x)$ . For all $x$ satisfying $f^{\prime}(f^{-1}(x))\ne 0$ ,

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

$\frac{dy}{dx}=\frac{d}{dx}(f^{-1}(x))=(f^{-1})^{\prime}(x)=\dfrac{1}{f^{\prime}(f^{-1}(x))}$

Alternatively, if $y=g(x)$ is the inverse of $f(x)$ , then

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

$g^{\prime}(x)=\dfrac{1}{f^{\prime}(g(x))}$

Example: Applying the Inverse Function Theorem, 1

Use the Inverse Function Theorem to find the derivative of $g(x)=\dfrac{x+2}{x}$ . Compare the resulting derivative to that obtained by differentiating the function directly.

Show Solution

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

f^{'} (x) = \frac{- 2}{(x - 1)^{2}}

$f^{\prime}(x)=\frac{-2}{(x-1)^2}$ and

f^{'} (g (x)) = \frac{- 2}{(g (x) - 1)^{2}} = \frac{- 2}{(\frac{x + 2}{x} - 1)^{2}} = - \frac{x^{2}}{2}

$f^{\prime}(g(x))=\frac{-2}{(g(x)-1)^2}=\frac{-2}{(\frac{x+2}{x}-1)^2}=-\frac{x^2}{2}$

Finally,

$g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}=-\frac{2}{x^2}$

We can verify that this is the correct derivative by applying the quotient rule to $g(x)$ to obtain

g^{'} (x) = - \frac{2}{x^{2}}

$g^{\prime}(x)=-\frac{2}{x^2}$

Watch the following video to see the worked solution to Example: Applying the Inverse Function Theorem, 1.

Closed Captioning and Transcript Information for Video

Try It

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

Show Solution

g^{'} (x) = - \frac{1}{(x + 2)^{2}}

$g^{\prime}(x)=-\frac{1}{(x+2)^2}$

Hint

Use the preceding example as a guide.

Example: Applying the Inverse Function Theorem, 2

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

Show Solution

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

f^{'} (x) = 3 x^{2}

$f^{\prime}(x)=3x^2$ and

f^{'} (g (x)) = 3 (\sqrt[3]{x})^{2} = 3 x^{2 / 3}

$f^{\prime}(g(x))=3(\sqrt[3]{x})^2=3x^{2/3}$

Finally,

g^{'} (x) = \frac{1}{3 x^{2 / 3}} = \frac{1}{3} x^{- 2 / 3}

$g^{\prime}(x)=\frac{1}{3x^{2/3}}=\frac{1}{3}x^{-2/3}$

Try It

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

Hint

Use the fact that $g(x)$ is the inverse of $f(x)=x^5$ .

Show Solution

$g(x)=\frac{1}{5}x^{−4/5}$

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.

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

$\frac{d}{dx}(x^{1/n})=\frac{1}{n}x^{(1/n)-1}$

Also, if $n$ is a positive integer and $m$ is an arbitrary integer, then

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

$\frac{d}{dx}(x^{m/n})=\frac{m}{n}x^{(m/n)-1}$

Proof

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

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

$f^{\prime}(x)=nx^{n-1}$ and

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

$f^{\prime}(g(x))=n(x^{1/n})^{n-1}=nx^{(n-1)/n}$ .

Finally,

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

$g^{\prime}(x)=\dfrac{1}{nx^{(n-1)/n}}=\frac{1}{n}x^{(1-n)/n}=\frac{1}{n}x^{(1/n)-1}$ .

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}

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

_{◼}

$_\blacksquare$

Example: Applying the Power Rule to a Rational Power

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

Show Solution

First find $\frac{dy}{dx}$ and evaluate it at $x=8$ . Since

\frac{d y}{d x} = \frac{2}{3} x^{- 1 / 3}

$\frac{dy}{dx}=\frac{2}{3}x^{-1/3}$ and

\frac{d y}{d x} |_{x = 8} = \frac{1}{3}

$\frac{dy}{dx}|_{x=8}=\frac{1}{3}$

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 $(8,4)$ . Substituting into the point-slope formula for a line and solving for $y$ , we obtain the tangent line

y = \frac{1}{3} x + \frac{4}{3}

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

Watch the following video to see the worked solution to Example: Applying the Power Rule to a Rational Power.

Closed Captioning and Transcript Information for Video

Try It

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

Hint

Rewrite as $s(t)=(2t+1)^{1/2}$ and use the chain rule.

Show Solution

$s^{\prime}(t)=(2t+1)^{−1/2}$

Module 3: Derivatives