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

17Calculus Derivatives - Inverse Trigonometric Functions

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This page discusses the derivatives of inverse trig functions.
You do not need to know the chain rule for the first part of this page, we discuss the basic derivatives first. Once you have learned the chain rule, you can come back here to work the practice problems.

If you want a full length lecture on inverse trig functions and their derivatives, we recommend the following video. The lecturer is one of our favorites and he is very good at explaining the inverse trig functions themselves and their derivatives using plenty of examples.
Note: We stop the video with about 20mins to go since he starts the next topic, integration. If you have already had integration, go ahead and watch the end of the video. It is good too.

Prof Leonard - Calculus 2 Lecture 6.5: Calculus of Inverse Trigonometric Functions [1hr-31min-10secs]

video by Prof Leonard

Basic Inverse Trig Derivatives (no chain rule)

You might expect the derivatives for inverse trig functions to be similar to derivatives for trig functions but they are not. They are very different. But, again, they appear in similar pairs.

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

 

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

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

 

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

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

 

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

Important Things To Notice

  1. Strangely enough, the derivatives of inverse trig functions do not contain any trig or inverse trig terms. We include some videos below showing the derivation of these equations explaining why this happens.

  2. Each pair of inverse trig derivatives are very closely related, even closer than with trig derivatives. Each pair is the same EXCEPT for a negative sign. So, for example, \( [\arccos(t)]' = -[\arcsin(t)]' \).

  3. In the case of the third pair, \( [\arcsec(t)]' \) and \( [\arccsc(t)]' \), the denominators contain an absolute value term, \( \abs{t} \), which is important. Do not leave off the absolute value signs unless you explicitly state that \( t \) is always positive. Keeping them is always the safe way to make sure you are correct. (Of course, you need to check with your instructor to see what they require.)

  4. Remember that the notation \(\sin^{-1}(t)\) actually means \(\arcsin(t)\), not \(1/\sin(t)\).

Before we go on, let's watch some videos showing how the derivatives above come about. It seems strange that when we take the derivative of a function involving inverse trig functions, there are no longer any trig or inverse trig terms. These next videos show how to get the derivatives. Watching these will help you understand trig and inverse trig in more depth.

PatrickJMT - The Derivative of Inverse Sine or y = arcsin(x) [7min-5secs]

video by PatrickJMT

MIP4U - Proof - The Derivative of f(x)=arccos(x): d/dx[arccos(x)] [3min-19secs]

video by MIP4U

PatrickJMT - Deriving the Derivative of Inverse Tangent or y = arctan (x) [6min-16secs]

video by PatrickJMT

MIP4U - Proof - The Derivative of f(x)=arccot(x): d/dx[arccot(x)] [4min-2secs]

video by MIP4U

MIP4U - Proof - The Derivative of f(x)=arcsec(x): d/dx[arcsec(x)] [4min-49secs]

video by MIP4U

MIP4U - Proof - The Derivative of f(x)=arccsc(x): d/dx[arccsc(x)] [4min-51secs]

video by MIP4U

Here are some practice problems. Calculate the derivative of these functions.

\(y=(1+x^2)\arctan(x)\)

Problem Statement

Calculate the derivative of \(y=(1+x^2)\arctan(x)\).

Solution

2152 video

video by PatrickJMT

close solution

Inverse Trig Derivatives Using The Chain Rule

As mentioned in the previous section, except for the simplest problems, you will need to use the chain rule for most inverse trig derivatives. Let's look at an example that parallels some of the examples in the previous section.

Evaluate \( [\arcsin(x^2)]' \)

\(\displaystyle{ [\arcsin(x^2)]' = \frac{2x}{\sqrt{1-x^4}} }\)

Problem Statement - Evaluate \( [\arcsin(x^2)]' \)
Solution - For this one, we let \( u = x^2 \). We chose \(x^2\) since we want \(u\) to be equal to everything inside the arcsine function.

\(\displaystyle{ \frac{d}{dx}[\arcsin(x^2)] }\)

let \( u = x^2 \)

\(\displaystyle{ \frac{d}{dx}[\arcsin(u)] }\)

Apply the chain rule.

\(\displaystyle{\frac{d}{du}[\arcsin(u)] \frac{d}{dx}[u]}\)

\(\displaystyle{\frac{d}{du}[\arcsin(u)] \frac{d}{dx}[x^2]}\)

\(\displaystyle{\frac{1}{\sqrt{1-u^2}} (2x)}\)

\(\displaystyle{\frac{1}{\sqrt{1-(x^2)^2}} (2x)}\)

\(\displaystyle{\frac{2x}{\sqrt{1-x^4}}}\)

Note -
Notice how, when we reverse substituted \( u = x^2 \), we had \( (x^2)^2 = x^4 \). This may be easily missed if you tried to do it in your head.

Final Answer

\(\displaystyle{ [\arcsin(x^2)]' = \frac{2x}{\sqrt{1-x^4}} }\)

close solution

Here are some practice problems. Unless otherwise instructed, calculate the derivative of these functions.

\(y=\sin^{-1}(2x+1)\)

Problem Statement

Calculate the derivative of \(y=\sin^{-1}(2x+1)\).

Solution

2155 video

video by Krista King Math

close solution
\(f(x) = (\sin^{-1}x)^2\)

Problem Statement

Calculate the derivative of \(f(x) = (\sin^{-1}x)^2\).

Solution

2156 video

video by PatrickJMT

close solution
\(y=\sqrt{\tan^{-1}x}\)

Problem Statement

Calculate the derivative of \(y=\sqrt{\tan^{-1}x}\).

Solution

2150 video

video by PatrickJMT

close solution
\(y=\tan^{-1}(\sqrt{x})\)

Problem Statement

Calculate the derivative of \(y=\tan^{-1}(\sqrt{x})\).

Solution

2151 video

video by PatrickJMT

close solution
\(y=e^{\sec^{-1}(t)}\)

Problem Statement

Calculate the derivative of \(y=e^{\sec^{-1}(t)}\).

Solution

2153 video

video by PatrickJMT

close solution
\(y=\sin^{-1}(x^2+1)\)

Problem Statement

Calculate the derivative of \(y=\sin^{-1}(x^2+1)\).

Solution

2154 video

video by PatrickJMT

close solution
\(g(t) = \cos^{-1}\sqrt{2t-1}\)

Problem Statement

Calculate the derivative of \(g(t) = \cos^{-1}\sqrt{2t-1}\).

Solution

2157 video

video by PatrickJMT

close solution
\(\displaystyle{y=\tan^{-1}(x/a) + \ln\sqrt{\frac{x-a}{x+a}}}\)

Problem Statement

Calculate the derivative of \(\displaystyle{y=\tan^{-1}(x/a) + \ln\sqrt{\frac{x-a}{x+a}}}\).

Solution

2158 video

video by PatrickJMT

close solution
\(y = \sec^{-1}\sqrt{1+x^2}\)

Problem Statement

Calculate the derivative of \(y = \sec^{-1}\sqrt{1+x^2}\).

Solution

2159 video

video by PatrickJMT

close solution
\(\displaystyle{ y = \sin^{-1}\left( \frac{\cos x}{1+\sin x} \right) }\)

Problem Statement

Calculate the derivative of \(\displaystyle{ y = \sin^{-1}\left( \frac{\cos x}{1+\sin x} \right) }\).

Solution

2160 video

video by PatrickJMT

close solution

You CAN Ace Calculus

Topics You Need To Understand For This Page

inverse functions

basic derivative rules

power rule

product rule

quotient rule

additional prerequisites for derivatives involving chain rule

chain rule

Related Topics and Links

inverse trig derivatives youtube playlist (check this playlist for additional practice problems not on this page)

Wikipedia - Derivatives of Inverse Trigonometric Functions

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