Inverse Trigonometric Functions and Derivatives

Inverse Sine

Let’s start with inverse sine. For a comprehensive review, check out our trig textbook. Here is the definition of the inverse sine function.

Sometimes instead of the -1 superscript, we name the inverse sine function arcsine, so the above information can be annotated as follows:

So, evaluating an inverse trig function is the same as asking what angle (i.e. $y$) did we plug into the sine function to get $x$. The restrictions on $y$ given above are there to make sure that we get a consistent answer out of the inverse sine. We know that there are in fact an infinite number of angles that will work and we want a consistent value when we work with inverse sine. Using the range of angles above gives all possible values of the sine function exactly once. If you’re not sure of that sketch out a unit circle and you’ll see that that range of angles (the $y$’s) will cover all possible values of sine.

Note as well that since $−1\leq\sin(y)\leq 1$ we also have $−1\leq x\leq 1$.

We have the following relationship between the inverse sine function and the sine function.

In other words sine and inverse sine are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,

Following the Inverse Function Theorem, we get

While this is correct, it is not a very useful formula. Let’s see if we can get a better formula. Let’s start by recalling the definition of the inverse sine function.

Using the first part of this definition the denominator in the derivative becomes,

To get a better idea of what we are looking at, let's look at a right triangle as if it were sitting in the unit circle in standard postion. For now, we may act as if it is sitting in the first quadrant.

What we really want to know is what $\cos(\theta)$ is, which can be determined by

As $\sin(\theta)=x=\dfrac{x}{1}=\dfrac{\text{opposite}}{\text{hypotenuse}}$, we can fill in some of the side lengths of this triangle.

We will need to know the adjacent side to find $\cos(\theta)$, which may find by using the Pythagorean Theorem.

Note that $\theta$ is restricted by $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$. Therefore, This triangle can only be viewed as if in quadrant 1 or 4. In either case, the adjacent side falls along the postive $x$-axis, meaning the adjacent side is positive. So we can finalize our triangle.

Finally, we can see that $$\cos(\theta)=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}.$$ Thus $\cos \left( {{{\sin }^{ - 1}}x} \right) = \cos \left( \theta \right)=\sqrt{1-x^2}$ and

Let's practice a few problems with the derivative of the inverse sine function.

Now try some on your own.

Inverse Cosine

Now recall some information about inverse cosine. Again, for a comprehensive review, check out our trig textbook. First, here is the definition of the inverse cosine function.

Once more, we could write arccosine instead of the superscript of -1 to indicate that we're dealing with the inverse of cosine, so the above information can also be written as

As with the inverse sine we’ve got a restriction on the angles, $y$, that we get out of the inverse cosine function. Again, if you’d like to verify this a quick sketch of a unit circle should convince you that this range will cover all possible values of cosine exactly once. Also, we also have $−1\leq x\leq 1$ because $−1\leq \cos(y)\leq 1$.

We have the following relationship between the inverse cosine function and the cosine function.

In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse cosine. As we did with the inverse sine function, let’s start with,

Using the Inverse Function Theorem again, we get

We could have left the negative in the denominator, but we move it here to have a slightly easier denominator to simplify.

And so, similar to the derivative for inverse sine, we can simplify $\sin(\cos^{-1}(x))$ and find that $\sin(\cos^{-1}(x))=\sqrt{1-x^2}$. We leave the details of this as an exercise.

Thus we can conclude,

Let's work with this derivative.

Now try some on your own.

Inverse Tangent

The last of these derivatives that we will cover in detail is that of the inverse tangent function. Check out a more comprehensive review of this function here. Recall the definition of the inverse tangent function.

We can annotate this using arctangent instead of the -1 superscript, so the above information is the same as writing

Again, we have a restriction on $y$, but notice that we can’t let $y$ be either of the two endpoints in the restriction above since tangent isn’t even defined at those two points. To convince yourself that this range will cover all possible values of tangent do a quick sketch of the tangent function and we can see that in this range we do indeed cover all possible values of tangent. Also, in this case there are no restrictions on $x$ because tangent can take on all possible values.

We have the following relationship between the inverse tangent function and the tangent function.

As these are inverses of each other, to find the derivative of the inverse tangent function we can start with

By the Inverse Function Theorem, we have

While we can use a right triangle as before to simplify $\sec^2\left(\tan^{-1}(x)\right)$, we can do so without. What we really need here is to use a Pythagorean Identity: $1+\tan^2(\theta)=\sec^2(\theta)$.

Firstly, let $\theta=\tan^{-1}(x)$; and so $\tan(\theta)=x$. So we see that

Therefore, we have the following derivative.

We also point out the alternate notion for the inverse tangent function.

Let's work with this derivative.

Now try some on your own.

Inverse Secant, Cosecant, and Cotangent

There are three more inverse trig functions but the three shown above the most common ones. Here are the definitions of the remaining inverse trigonometric functions.

We point out that some authors use differe restrictions on the values of $y$ for the inverse secant and inverse cosecant functions. Specifically, they use $0\le y \lt \frac{\pi}{2} \hspace{0.75em}\text{or}\hspace{0.75em} \pi \le y\lt \frac{3\pi}{2}$ for inverse secant and $-\pi\lt y \le -\frac{\pi}{2} \hspace{0.75em}\text{or}\hspace{0.75em} 0 \lt y\le \frac{\pi}{2}$. For this course, we use the restrictions given above.

These three functions also have similar alternate notations as the previous three functions.

Formulas for the derivatives of these three could be derived by a similar process as we did with the inverse sine, cosine, and tangent functions. We give those derivatives and recap the derivatives of all inverse trigonometric functions.

Practice Problems

For problems 1 – 7 differentiate the given function.

1. $T\left( z \right) = 2\cos \left( z \right) + 6{\cos ^{ - 1}}\left( z \right)$
Click for solution


Not much to do here other than take the derivative using the formulas.

$$T'\left( z \right) = - 2\sin \left( z \right) - \frac{6}{{\sqrt {1 - {z^2}} }}$$


2. $g\left( t \right) = {\csc ^{ - 1}}\left( t \right) - 4{\cot ^{ - 1}}\left( t \right)$
Click for solution


Not much to do here other than take the derivative using the formulas.

$$g'\left( t \right) = - \frac{1}{{\left|t\right|\sqrt {{t^2} - 1} }} + \frac{4}{{{t^2} + 1}}$$


3. $y = 5{x^6} - {\sec ^{ - 1}}\left( x \right)$
Click for solution


Not much to do here other than take the derivative using the formulas.

$$\frac{{dy}}{{dx}} = 30{x^5} - \frac{1}{{\left|x\right|\sqrt {{x^2} - 1} }}$$


4. $f\left( w \right) = \sin \left( w \right) + {w^2}{\tan ^{ - 1}}\left( w \right)$
Click for solution


Other than following the derivative formulas, we will need to use the product rule with the second term of the function.

$$f'\left( w \right) = \cos \left( w \right) + 2w{{\tan }^{ - 1}}\left( w \right) + \frac{{{w^2}}}{{1 + {w^2}}}$$


5. $\displaystyle h\left( x \right) = \frac{{{{\sin }^{ - 1}}\left( x \right)}}{{1 + x}}$
Click for solution


Note that the function is a quotient, thus we will need to use the quotient rule overall. We will also simplify the complex fraction that results from the quotient rule and derivative of the inverse sine function.

$$h'\left( x \right) = \frac{{\frac{{1 + x}}{{\sqrt {1 - {x^2}} }} - {{\sin }^{ - 1}}\left( x \right)}}{{{{\left( {1 + x} \right)}^2}}} = \frac{{1 + x - \sqrt {1 - {x^2}} {{\sin }^{ - 1}}\left( x \right)}}{{\sqrt {1 - {x^2}} {{\left( {1 + x} \right)}^2}}}$$


6. $u\left( t \right) = {\tan ^{ - 1}}\left( {3t - 1} \right)$
Click for solution


For this problem, we will use the chian rule where the outside function is the inverse tangent and the inside function is the stuff inside of the inverse tangent.

$$u'\left( t \right) = \frac{3}{{{{\left( {3t - 1} \right)}^2} + 1}}$$


7. $g\left( x \right) = {\left( {\ln \left( {{x^2} + 1} \right) - {{\tan }^{ - 1}}\left( {6x} \right)} \right)^{10}}$
Click for solution


This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear.

$$\begin{align*} g'\left( x \right) &= 10{\left( {\ln \left( {{x^2} + 1} \right) - {{\tan }^{ - 1}}\left( {6x} \right)} \right)^9}\frac{d}{{dx}}\left( {\ln \left( {{x^2} + 1} \right) - {{\tan }^{ - 1}}\left( {6x} \right)} \right) \\ &= 10{{\left( {\ln \left( {{x^2} + 1} \right) - {{\tan }^{ - 1}}\left( {6x} \right)} \right)}^9}\left( {\frac{{2x}}{{{x^2} + 1}} - \frac{6}{{36{x^2} + 1}}} \right) \end{align*}$$

Assignment Problems

For problems 1 – 15 differentiate the given function.

$f\left( x \right) = \sin \left( x \right) + 9{\sin ^{ - 1}}\left( x \right)$


$C\left( t \right) = 5{\sin ^{ - 1}}\left( t \right) - {\cos ^{ - 1}}\left( t \right)$


$g\left( z \right) = {\tan ^{ - 1}}\left( z \right) + 4{\cos ^{ - 1}}\left( z \right)$


$h\left( t \right) = {\sec ^{ - 1}}\left( t \right) - {t^3}{\cos ^{ - 1}}\left( t \right)$


$f\left( w \right) = \left( {w - {w^2}} \right){\sin ^{ - 1}}\left( w \right)$


$y = \left( {x - {{\cot }^{ - 1}}\left( x \right)} \right)\left( {1 + {{\csc }^{ - 1}}\left( x \right)} \right)$


$\displaystyle Q\left( z \right) = \frac{{z + 1}}{{{{\tan }^{ - 1}}\left( z \right)}}$


$\displaystyle A\left( t \right) = \frac{{1 + {{\sin }^{ - 1}}\left( t \right)}}{{1 - {{\cos }^{ - 1}}\left( t \right)}}$


$f\left( v \right) = {\tan ^{ - 1}}\left( {3 - 2v} \right)$


$h\left( t \right) = {\sin ^{ - 1}}\left( {9t} \right)$


$R(x)=\ln\left(\cos^{-1}(x)\right)$


$b(z)=\sqrt{3z+2}\arccos(4z)$


$k(w)=\tan^{-1}\left(\sec^{2}(w)\right)$


$v(t)=\arcsin\left(t\ln(t)\right)$


$g(x)=\arctan\left(\dfrac{e^t}{t+1}\right)$




1. Find the tangent line to $f(x)=5\tan^{-1}\left(\dfrac{x}{2}\right)$ at $x=2$.
2. Find the tangent line to $h(x)=\dfrac{\arcsin(3x)}{x}$ at $x=-\dfrac{\sqrt{3}}{6}$.

Note: The following content is adapted from OpenStax and is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction.

Differentiate each function.

$y = \arcsin(x^2)$


$y=\arccos(\sqrt{x})$


$y= \sqrt{\csc^{-1}(x)}$


$y = \cot^{-1}\left(\sqrt{4-x^2}\right)$


$ y = \left(1+\arctan(x)\right)^3$


$y = \arccos(2x)\cdot\arcsin(2x)$