Derivatives of Inverse Functions: Learn It 1

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Derivatives of Inverse Functions: Learn It 1

Find the derivative of an inverse function
Identify the derivatives for inverse trig functions like arcsine, arccosine, and arctangent

Derivatives of Various Inverse Functions

Understanding the relationship between the derivatives of a function and its inverse is crucial when extending the application of derivatives to inverse functions without needing the limit definition every time. We start by exploring how the derivative of a function and its inverse are interlinked.

If a function $f(x)$ is both invertible and differentiable, it logically follows that its inverse $f^{−1}(x)$ should also be differentiable.

Consider the function $f(x)$ and its inverse $f^{−1}(x)$ as depicted in Figure 1, where the point on the graph of $(a,f ^{−}1 (a))$ on the graph of $f^{−1}(x)$ and the point $(f^{−1}(a), a)$ on the graph of $f(x)$ demonstrate this relationship.

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.

At these points, the tangent lines of the function and its inverse have reciprocal slopes due to their symmetrical relationship about the line $y=x$ . Specifically, if the slope of the tangent at $a$ on $f(x)$ is $f ^{\prime} (f^{ −1} (a))= \frac{q}{p} $ , then the slope at $f^{−1}(a)$ on $f^{−1}(x)$ must be $\frac{q}{p}$ . This is illustrated in the reciprocal nature of their slopes, confirming that:

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

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

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))}

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

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.

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 this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.7 Derivatives of Inverse Functions (edited)” here (opens in new window).

Find the derivative of

g (x) = \sqrt[5]{x}

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

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

$g(x)=\frac{1}{5}x^{−4/5}$

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}

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

$_\blacksquare$

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

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

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

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.7 Derivatives of Inverse Functions (edited)” here (opens in new window).