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.

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.

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.

• Example
• Answer
• Solution