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 (affiliate links) | ||
---|---|---|
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.
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.
video by PatrickJMT |
---|
video by MIP4U |
---|
video by PatrickJMT |
---|
video by MIP4U |
---|
video by MIP4U |
---|
video by MIP4U |
---|
Here are some practice problems.
Unless otherwise instructed, calculate the derivative of these functions.
\(y=(1+x^2)\arctan(x)\)
Problem Statement
Calculate the derivative of this function, giving your answer in completely factored form. \(y=(1+x^2)\arctan(x)\)
Solution
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
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 \( [\arcsin(x^2)]' \)
Final Answer
\(\displaystyle{ [\arcsin(x^2)]' = \frac{2x}{\sqrt{1-x^4}} }\) |
---|
Problem Statement
Evaluate \( [\arcsin(x^2)]' \)
Solution
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
\(\displaystyle{ [\arcsin(x^2)]' = \frac{2x}{\sqrt{1-x^4}} }\) |
---|
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 this function, giving your answer in completely factored form. \( y = \sin^{-1}(2x+1) \)
Solution
video by Krista King Math |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\( f(x) = (\sin^{-1}x)^2 \)
Problem Statement
Calculate the derivative of this function, giving your answer in completely factored form. \( f(x) = (\sin^{-1}x)^2 \)
Solution
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\( y = \sqrt{\tan^{-1}x} \)
Problem Statement
Calculate the derivative of this function, giving your answer in completely factored form. \( y = \sqrt{\tan^{-1}x} \)
Solution
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\( y = \tan^{-1}(\sqrt{x}) \)
Problem Statement
Calculate the derivative of this function, giving your answer in completely factored form. \( y = \tan^{-1}(\sqrt{x}) \)
Solution
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\(y = e^{\sec^{-1}(t)} \)
Problem Statement
Calculate the derivative of this function, giving your answer in completely factored form. \(y = e^{\sec^{-1}(t)} \)
Solution
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\(y=\sin^{-1}(x^2+1)\)
Problem Statement
Calculate the derivative of this function, giving your answer in completely factored form. \(y=\sin^{-1}(x^2+1)\)
Solution
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\( g(t) = \cos^{-1}\sqrt{2t-1} \)
Problem Statement
Calculate the derivative of this function, giving your answer in completely factored form. \( g(t) = \cos^{-1}\sqrt{2t-1} \)
Solution
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\(\displaystyle{ y = \tan^{-1}(x/a) + \ln\sqrt{\frac{x-a}{x+a}} }\)
Problem Statement
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
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\(y = \sec^{-1}\sqrt{1+x^2}\)
Problem Statement
Calculate the derivative of this function, giving your answer in completely factored form. \(y = \sec^{-1}\sqrt{1+x^2}\)
Solution
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
\(\displaystyle{ y = \sin^{-1}\left( \frac{\cos x}{1+\sin x} \right) }\)
Problem Statement
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
video by PatrickJMT |
---|
Log in to rate this practice problem and to see it's current rating. |
---|
Really UNDERSTAND Calculus
Log in to rate this page and to see it's current rating.
inverse functions |
additional prerequisites for derivatives involving chain rule |
---|
inverse trig derivatives youtube playlist (check this playlist for additional practice problems not on this page) |
To bookmark this page and practice problems, log in to your account or set up a free account.
Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
| |
I recently started a Patreon account to help defray the expenses associated with this site. To keep this site free, please consider supporting me. |
---|
Support 17Calculus on Patreon |