# Derivatives of Inverse Trig Functions

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Objectives
• Know the derivatives of arcsin(x), arccos(x), arctan(x)
• See (but perhaps not memorize) the derivatives of arccot(x), arcsec(x), and arccsc(x)
• Understand similarities and relationships among these derivatives to aide memorization
• Learn a $u$ substitution formula to take the derivative of any inverse trig function when the input is not merely $x$ (previewing the Chain Rule)
Lesson Description

We will learn how to take the derivative of the inverse trig functions we studied in trigonometry. While this lesson is very commonly tested, it is almost entirely a matter of memorization. You will only see this again infrequently in the future, but if you do, many teachers expect you to remember it.

Practice Problems

Practice problems and worksheet coming soon!

## Unpopular Functions, Unpopular Derivatives

Let's face it - trig is already not everyone's favorite subject. More students groan than smile when they realize that Calculus asks you to remember some fundamental trig concepts and to often work with sine and cosine functions. So when it comes to inverse trig functions, which, as far as functions and function behavior goes, are awkward and hard to remember, I'm used to extra sighs, eye rolls, and yawns in any session that includes them.Fortunately, we won't see these too often in Calculus, but we are expected to recognize them when we do. For better or worse, your usual task here is, simply stated, to memorize three formulas, two of which are practically identical.

## Memorize These Formulas

Here's the real deal: you need to memorize the formula for the derivatives of $\sin^{-1}(x)$ and $\tan^{-1}(x)$. The third derivative you'll need is that of $\cos^{-1}(x)$. Together, these three comprise the knowledge that both the AP exams and college courses expect you to have.
Core Inverse Trig Function DerivativesThe following two derivatives should be committed to memory:$$\frac{d}{dx} \, \sin^{-1} (x) = \frac{1}{\sqrt{1-x^2}}$$$$\frac{d}{dx} \, \tan^{-1} (x) = \frac{1}{1+x^2}$$Additionally, the derivative of the inverse cosine function is required knowledge, but you should make note of its close resemblance to the inverse sine function's derivative.$$\frac{d}{dx} \, \cos^{-1} (x) = -\frac{1}{\sqrt{1-x^2}}$$
That's all there is to say about the baseline knowledge here - know the formulas and recite them when asked.

## Mr. Math Makes It Mean

Often times, when studying these specific derivatives, teachers expect you to be able to compute derivatives using the chain rule ». Whether you've covered this concept specifically, this is a feasible task if you modify our formulas to adjust for inputs into inverse trig functions other than $x$.
Inverse Trig Derivatives for Inputs Other Than xIf you need to take the derivative of an inverse trig function where the input to the function is not $x$, but rather some expression $u$, modify the derivative formulas by
1. replacing $x$ with $u$ in the formulas
2. multiplying the result by the derivative of $u$
In symbols, this definition says$$\frac{d}{dx} \, \sin^{-1} (u) = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$$$\frac{d}{dx} \, \cos^{-1} (u) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$$$\frac{d}{dx} \, \tan^{-1} (u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$$
Let's try an example.

Example 1Find the derivative of $\tan^{-1}(x)$, and then find the derivative of $\tan^{-1} \left( x^2 \right)$.$\blacktriangleright$ The first ask is verbatim knowledge we are going to be asked to know and memorize.$$\frac{d}{dx} \, \tan^{-1} (x) = \frac{1}{1+x^2}$$The second part of this problem requires this chain rule approach, using the $u$ formula.$$\frac{d}{dx} \, \tan^{-1} (u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$$\begin{align} \Rightarrow \frac{d}{dx} \, \tan^{-1} \left( x^2 \right) & = \frac{1}{1+\left(x^3\right)^2} \cdot \frac{d}{dx} \, x^3 \\ & = \frac{3x^2}{1+x^6} \end{align}
You Should Know While the process of taking derivatives of compositions of functions is technically the Chain Rule, inverse trig derivatives are often taught in this format, and students are expected to be able to compute these types of derivatives regardless of whether the Chain Rule is yet to be covered in class. As long as you can follow what's being done, it doesn't matter whether we call it Chain Rule or just think of it as a formula.

## Unusual Suspects

If your teacher really insists, you may have to memorize the formula for the derivative of $\sec^{-1}(x)$. It's not on the AP syllabus, and because of that, most teachers won't bother with it. In fact, many college profs don't either.
Derivative of Secant InverseThe derivative of the inverse secant function is as follows.$$\frac{d}{dx} \, \sec^{-1} (x) = \frac{1}{\left( |x| \sqrt{x^2 - 1} \right)}$$For any expression $u$, the derivative is$$\frac{d}{dx} \, \sec^{-1} (u) = \frac{1}{\left( |u| \sqrt{u^2 - 1} \right)} \, \cdot \frac{du}{dx}$$
Just like the sine-cosine relationship, the "co" functions have the same derivatives as their namesake functions, but with a negative sign. So while it is also unusual to be asked to know about inverse cotangent and inverse cosecant, their derivatives aren't any more difficult than knowing the inverse tangent and inverse secant derivatives - the former you are asked to know for sure, and the latter likely not, as discussed.
Derivative of Inverse "co" FunctionsThe derivatives of the three inverse co-functions, including inverse cosine which we already looked at, are as follows.$$\frac{d}{dx} \, \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}$$$$\frac{d}{dx} \, \cot^{-1} (x) = -\frac{1}{1+x^2}$$$$\frac{d}{dx} \, \csc^{-1} (x) = -\frac{1}{ |x| \sqrt{x^2 - 1}}$$
Compare these with the derivatives of the "non-co" trig functions.$$\frac{d}{dx} \, \sin^{-1} (x) = \frac{1}{\sqrt{1-x^2}}$$$$\frac{d}{dx} \, \tan^{-1} (x) = \frac{1}{1+x^2}$$$$\frac{d}{dx} \, \sec^{-1} (x) = \frac{1}{|x| \sqrt{x^2 - 1}}$$
Pro Tip The negative "co" trick is not new. This pattern un-coincidentally appears in regular trig function derivatives as well, but while the reason this happens with regular trig functions was worth understanding, it's not worth trying to make any meaningful connection why this happens here, if you ask me.For these derivatives, consider the inverse cosine function mandatory, the inverse cotangent function fair game, and the inverse cosecant function sharing the same level of optionality as the secant (that is, if you need to know one, you should know the other).
Finally, as with the other inverse trig function derivatives, we are commonly asked to learn the formulas in "chain rule" form, whether or not we learned the Chain Rule yet. For any expression $u$ which is itself some $x$ expression, we have$$\frac{d}{dx} \, \cos^{-1}(u) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$$$\frac{d}{dx} \, \cot^{-1} (u) = -\frac{1}{1+u^2} \cdot \frac{du}{dx}$$$$\frac{d}{dx} \, \csc^{-1} (u) = -\frac{1}{ |u| \sqrt{u^2 - 1} } \cdot \frac{du}{dx}$$

## Put It To The Test

Compute each of the following derivatives.

Example 2$$\frac{d}{dx} \, \cos^{-1}(x)$$
Show solution
$\blacktriangleright$ This derivative is just a copy / paste of the formula. As a tip for memorizing this one, recall how similar it is to the derivative of $\sin^{-1}(x)$.$$\frac{d}{dx} \, \cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}$$

Example 3$$\frac{d}{dx} \, \sin^{-1}(3x)$$
Show solution
$\blacktriangleright$ Using the $u$ formula, we can compute this derivative by multiplying by the derivative of $3x$. Just make sure to use parenthesis to avoid formula mistakes.$$\frac{d}{dx} \, \sin^{-1}(3x) = \frac{1}{\sqrt{1-(3x)^2}} \cdot 3$$$$=\frac{3}{\sqrt{1-9x^2}}$$

Example 4$$\frac{d}{dx} \, \cot^{-1}(\sin(x))$$
Show solution
$\blacktriangleright$ Proceed with the $u$ type formula, and clean up the result. It's ok that you'll have trig functions in your answer.$$\frac{d}{dx} \, \cot^{-1}(\sin(x)) = \frac{1}{1 + (\sin(x))^2} \cdot \frac{d}{dx} \, \sin(x)$$$$=\frac{\cos(x)}{1 + \sin^2 (x)}$$

Example 5$$\frac{d}{dx} \, \cos^{-1}(e^{2x})$$
Show solution
$\blacktriangleright$ Using the formula,$$\frac{d}{dx} \, \cos^{-1}(e^{2x}) = -\frac{1}{\sqrt{1-\left(e^{2x}\right)^2}} \cdot \frac{d}{dx} \, e^{2x}$$$$=-\frac{2e^{2x}}{\sqrt{1-e^{4x}}}$$

Example 6$$\frac{d}{dx} \, \sec^{-1}\left( \frac{x}{2} \right)$$
Show solution
This is a challenging problem in that, as mentioned, many of you won't be exposed to this formula, never mind expected to know this formula.$$\frac{d}{dx} \, \sec^{-1}\left( \frac{x}{2} \right) = \frac{1}{\bigg| \frac{x}{2} \,\, \bigg| \cdot \sqrt{\left(\frac{x}{2}\right)^2-1}} \cdot \frac{d}{dx} \, \frac{x}{2}$$$$=\frac{1}{|x| \sqrt{\frac{x^2}{4} - 1}}$$

Lesson Takeaways
• Recall what inverse trig functions are, at least at a level of recognition
• Memorize the derivatives of $\sin^{-1}(x)$ and $\tan^{-1}(x)$
• Memorize the derivative of $\cos^{-1}(x)$ as an extension of $\sin^{-1}(x)$
• Figure out whether your teacher is going to cover all six inverse trig functions even though the AP syllabus focuses on three
• Master a light version of the Chain Rule with these formulas by practicing derivatives of inverse trig functions with $x$ expression inputs, not just $x$
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