\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

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There are times when we need to know the derivative of an inverse function but it is not possible to calculate the actual inverse function.
For a full lecture on this topic, we recommend this video.

Prof Leonard - Calculus 2 Lecture 6.2: Derivatives of Inverse Functions [44mins-8secs]

video by Prof Leonard

When we are not able to calculate the inverse of a function but we need to know the derivative of the inverse at a point, we can use implicit differentiation to come up with a formula to do the calculation.

We need to be careful though to make sure that the function actually has an inverse. In order for a function to have an inverse, it must be one-to-one. One way to make sure it is one-to-one is to check that the function is always either increasing or decreasing in the interval we are concerned about. Of course, using calculus, we can take the derivative and find the intervals where the function is increasing and decreasing. Let's look at an example.

Example 1

built with GeoGebra

Does the function \( f(x) = x^3 + 7x + 2 \) have an inverse?
Let's take the derivative and see if we can determine anything from it.
\( f'(x) = 3x^2 + 7 \)
Notice that the derivative is always positive for all x since \(3x^2\) is positive and adding a positive 7 keeps the result positive. So, since the function is always increasing, it is one-to-one on it's entire domain and therefore has an inverse. To help you visualise this, here is the graph.

So, what do we do when have a function, we know it has an inverse but we can't find the inverse in order to take the derivative? Well, we can use implicit differentiation to find an expression for the derivative at a point as follows.

\( \begin{array}{rcl} y & = & f(x) \\ f^{-1}(y) & = & f^{-1}(f(x)) \\ f^{-1}(y) & = & x \\ \displaystyle{ \frac{d}{dx} } [ f^{-1}(y) ] & = & \displaystyle{ \frac{d}{dx} } [x] \\ (f^{-1}(y))' \displaystyle{ \frac{dy}{dx} } & = & 1 \\ (f^{-1}(y))' & = & \displaystyle{ \frac{1}{dy/dx} } \\ (f^{-1}(y))' & = & \displaystyle{ \frac{1}{f'(x)} } \\ \end{array} \)

Now we have an expression for the derivative of an inverse function at the point \((x_1,y_1)\) on \(f(x)\) which is the point \((y_1,x_1)\) on \(f^{-1}(x)\). \[ (f^{-1}(y_1))' = \frac{1}{f'(x_1)} \] The tick marks used to write the derivatives above hide important details of this equation. To more clearly show what is going on, we can write the equation like this. \[ \frac{d}{dy} [f^{-1}(y)] = \frac{1}{dy/dx} \] Notice that the derivative on the left is with respect to y and the derivative on the right is with respect to x.

This may still be a bit confusing, so let's look at it another way. If we have a function \(f(x)\) with an inverse \(g(x)\), then \[ g'(x) = \frac{1}{f'(g(x))} \] at the point \((x,g(x))\).

So how do we use all these equations to solve a problem? Let's go through an example to demonstrate how to keep everything straight.

Example 2

For the function \(f(x)=x^2; x \gt 0\), calculate \((f^{-1}(9))'\).
For this example, it is possible to find the inverse for \(x \gt 0 \) and we will do that later to check our answer. But let's use the above equations to get an answer first. We will build a table to keep track of everything. The first row shows information about \(f(x)\) and the second row shows information about the inverse of \(f(x)\).

\( f(x) = x^2 \)

\( f'(x) = 2x \)

\( f'(g(9)) \)

\(f^{-1}(x) = g(x)\)

\( (x, g(x) ) = (9,3)\)

\( g'(9) \)

We have a partially filled in table above with some basic information. We are given \(f(x)\) and the point \(x=9\) on the inverse function. We changed the notation a bit and wrote \(f^{-1}(x) = g(x)\). Using \(g(x)\) makes the notation simpler and easier to use. In the above table we took the derivative of \(f(x)\) and calculated the point \((9,3)\). To get the 3, we note that we are given \(x=9\) on \(g(x)\) and we know that since \(f(x)\) and \(g(x)\) are inverses, then \(f(x) = 9\). This means that \(x^2=9 \to x=\pm 3\) and since the problem stated that \(x \gt 0\), we know that \(x=3\). So that gives us the point \((9,3)\).

Now we can calculate \( f'(g(x)) \). From the table, \(g(9)=3\), so \(f'(g(9)) = f'(3)\) giving us \(f'(g(9)) = 6\). So we will fill in that spot in the table and the result in the last row and last column, which comes from \( g'(9) = 1/f'(g(9)) \).

\( f(x) = x^2 \)

\( f'(x) = 2x \)

\( f'(g(9)) = 6 \)

\( g(x) \)

\( (x, g(x) ) = (9,3)\)

\( g'(9) = 1/6 \)

Let's check our answer by directly calculating the derivative of the inverse function.
\( \begin{array}{rcl} g(x) & = & \sqrt{x} \\ g'(x) & = & \displaystyle{ \frac{1}{2x^{1/2}} } \\ g'(9) & = & \displaystyle{ \frac{1}{(2)9^{1/2}} = \frac{1}{6} } \text{✔} \end{array} \)

Final Answer

\( (f^{-1}(9))' = g'(9) = 1/6 \)

The key to working these problems correctly is to keep track of the x and y values and knowing what to use when. We think that the using a table, like we did in the second example above, helps.

Before you try some on your own, watch this quick video clip with a simplified explanation.

Michael Griffis - Calculus 1-The Derivative of an Inverse Function [4mins-34secs]

video by Michael Griffis

Okay, time for some practice problems.

Practice

If \( f(x)=x^3+e^x \) and \( f^{-1}(1)=0 \), calculate the derivative of \(f^{-1}\) at \(x=1\). Make sure you check that this function has an inverse.

Problem Statement

If \( f(x)=x^3+e^x \) and \( f^{-1}(1)=0 \), calculate the derivative of \(f^{-1}\) at \(x=1\). Make sure you check that this function has an inverse.

Solution

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video by PatrickJMT

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If \( f(x) = \cos x + 3x \) and \( f^{-1}(3\pi/2) = \pi/2 \), calculate the derivative of \( f^{-1} \) at \( x=3\pi/2 \). Make sure to check that \( f(x) \) has an inverse.

Problem Statement

If \( f(x) = \cos x + 3x \) and \( f^{-1}(3\pi/2) = \pi/2 \), calculate the derivative of \( f^{-1} \) at \( x=3\pi/2 \). Make sure to check that \( f(x) \) has an inverse.

Solution

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For \( f(x)=2x^3+3x^2+7x+4 \), calculate \( (f^{-1})'(4) \).

Problem Statement

For \( f(x)=2x^3+3x^2+7x+4 \), calculate \( (f^{-1})'(4) \).

Solution

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video by VinTeachesMath

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For \( f(x)=x^3+2x-1\), calculate \( (f^{-1})'(2) \).

Problem Statement

For \( f(x)=x^3+2x-1\), calculate \( (f^{-1})'(2) \).

Solution

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For \(\displaystyle{ f(x)=\frac{1}{27}(x^5+2x^3) }\), calculate \( (f^{-1})'(-11) \).

Problem Statement

For \(\displaystyle{ f(x)=\frac{1}{27}(x^5+2x^3) }\), calculate \( (f^{-1})'(-11) \).

Solution

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For \( f(x) = x^3-4/x \), calculate \( (f^{-1})'(6) \).

Problem Statement

For \( f(x) = x^3-4/x \), calculate \( (f^{-1})'(6) \).

Solution

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Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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