17Calculus Derivatives - Inverse Trigonometric Functions
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.
Recommended Books on Amazon | ||
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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 |
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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 |
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MIP4U - Proof - The Derivative of f(x)=arccos(x): d/dx[arccos(x)] [3min-19secs]
video by MIP4U |
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PatrickJMT - Deriving the Derivative of Inverse Tangent or y = arctan (x) [6min-16secs]
video by PatrickJMT |
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MIP4U - Proof - The Derivative of f(x)=arccot(x): d/dx[arccot(x)] [4min-2secs]
video by MIP4U |
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MIP4U - Proof - The Derivative of f(x)=arcsec(x): d/dx[arcsec(x)] [4min-49secs]
video by MIP4U |
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MIP4U - Proof - The Derivative of f(x)=arccsc(x): d/dx[arccsc(x)] [4min-51secs]
video by MIP4U |
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Here are some practice problems.
Unless otherwise instructed, calculate the derivative of these functions.
\(y=(1+x^2)\arctan(x)\)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \(y=(1+x^2)\arctan(x)\)
Solution |
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2152 video
video by PatrickJMT |
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close solution
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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.
Evaluate \( [\arcsin(x^2)]' \)
\(\displaystyle{ [\arcsin(x^2)]' = \frac{2x}{\sqrt{1-x^4}} }\)
Evaluate \( [\arcsin(x^2)]' \)
Solution |
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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 |
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\(\displaystyle{ [\arcsin(x^2)]' = \frac{2x}{\sqrt{1-x^4}} }\) |
Here are some practice problems.
Unless otherwise instructed, calculate the derivative of these functions.
\( y = \sin^{-1}(2x+1) \)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \( y = \sin^{-1}(2x+1) \)
Solution |
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2155 video
video by Krista King Math |
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\( f(x) = (\sin^{-1}x)^2 \)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \( f(x) = (\sin^{-1}x)^2 \)
Solution |
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2156 video
video by PatrickJMT |
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close solution
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\( y = \sqrt{\tan^{-1}x} \)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \( y = \sqrt{\tan^{-1}x} \)
Solution |
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2150 video
video by PatrickJMT |
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\( y = \tan^{-1}(\sqrt{x}) \)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \( y = \tan^{-1}(\sqrt{x}) \)
Solution |
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2151 video
video by PatrickJMT |
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\(y = e^{\sec^{-1}(t)} \)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \(y = e^{\sec^{-1}(t)} \)
Solution |
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2153 video
video by PatrickJMT |
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\(y=\sin^{-1}(x^2+1)\)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \(y=\sin^{-1}(x^2+1)\)
Solution |
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2154 video
video by PatrickJMT |
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\( g(t) = \cos^{-1}\sqrt{2t-1} \)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \( g(t) = \cos^{-1}\sqrt{2t-1} \)
Solution |
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2157 video
video by PatrickJMT |
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\(\displaystyle{ y = \tan^{-1}(x/a) + \ln\sqrt{\frac{x-a}{x+a}} }\)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \(\displaystyle{ y = \tan^{-1}(x/a) + \ln\sqrt{\frac{x-a}{x+a}} }\)
Solution |
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2158 video
video by PatrickJMT |
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\(y = \sec^{-1}\sqrt{1+x^2}\)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \(y = \sec^{-1}\sqrt{1+x^2}\)
Solution |
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2159 video
video by PatrickJMT |
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\(\displaystyle{ y = \sin^{-1}\left( \frac{\cos x}{1+\sin x} \right) }\)
Problem Statement |
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Calculate the derivative of this function, giving your answer in completely factored form. \(\displaystyle{ y = \sin^{-1}\left( \frac{\cos x}{1+\sin x} \right) }\)
Solution |
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2160 video
video by PatrickJMT |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
inverse functions |
additional prerequisites for derivatives involving chain rule |
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Related Topics and Links
inverse trig derivatives youtube playlist (check this playlist for additional practice problems not on this page) |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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Practice Instructions
Unless otherwise instructed, calculate the derivative of these functions.
