Derivatives of Inverse Trigonometric Functions

Notation of Inverse Trigonometric Functions

Before starting, we shall talk briefly about the notation used for inverse trigonometric functions, which are also known as the arcus functions.

It is worth noting that the minus one is not an exponent. It is used to state that the function is an inverse, unlike \( \sin^{2}{x},\) where the two is an exponent telling us that the output of the sine function is to be squared.

Formulas for the Derivatives of Inverse Trigonometric Functions

With the notation clarified, let's take a look at the formulas for the derivatives of the six inverse trigonometric functions.

Method for finding the Derivatives of Inverse Trigonometric Functions

Just like with the derivatives of other functions, the method for finding the derivative of an inverse trigonometric function depends on the function. Let's see how this is done.

1. Identify which differentiation rule(s) is (are) relevant.

2. Use the above differentiation rule(s).

3. Write the derivative(s) of the inverse trigonometric function(s), as well as any other functions involved in the calculation.

As usual, these steps are better understood looking at examples. Let's jump into the next section!

Examples of the Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions can be used along with other differentiation rules like the chain rule, the product rule, and the quotient rule. Let's take a look at an example of each case!

How about the product rule?

Proofs of the Derivatives of Inverse Trigonometric Functions

You might have noticed that the derivatives of trigonometric functions involve other trigonometric functions but the derivatives of inverse trigonometric functions do not. To better understand why this happens, we will take a look at the proof of the derivative of each inverse trigonometric function.

Derivative of Inverse Sine

Let's begin by recalling that the inverse sine function is related to the sine function by the fact that they are each other's inverses. This means that

$$y=\arcsin{x} \mbox{ is true if and only if } \sin{y}=x.$$

Next, differentiate both sides of \( \sin{y}=x,\) so

$$\frac{\mathrm{d}}{\mathrm{d}x}\sin{y}=\frac{\mathrm{d}}{\mathrm{d}x} x.$$

The derivative of the sine function is the cosine function, but since \( y\) is a function of \( x, \) you have to use the chain rule on the left-hand side of the equation. The right-hand side of the equation is the derivative of \(x,\) so it is just 1. This will give you

$$(\cos{y})\frac{\mathrm{d}y}{\mathrm{d}x} =1,$$

where you can use the trigonometric Pythagorean identity,

$$\sin^2{\theta}+\cos^2{\theta}=1,$$ to write the cosine in terms of the sine. Doing this gives you

$$\left(\sqrt{1-\sin^2{y}}\right)\frac{\mathrm{d}y}{\mathrm{d}x} =1.$$

Next, substitute back \( \sin{y}=x \) to get

$$\left(\sqrt{1-x^2}\right)\frac{\mathrm{d}y}{\mathrm{d}x} =1.$$

Then isolate the derivative of \( y \),

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\sqrt{1-x^2}},$$

which is the formula for differentiating the inverse sine function

$$\frac{\mathrm{d}}{\mathrm{d}x} \arcsin{x}=\frac{1}{\sqrt{1-x^2}}.$$

You can try finding the derivatives of the inverse cosine, inverse tangent, and inverse cotangent in a similar way.

Derivative of Inverse Cosecant

Since you already found the derivative of the inverse sine function, so you can use this to your advantage! Since the cosecant function is the reciprocal of the sine function, you can write the identity

$$y=\mathrm{arccsc}{\,x}=\arcsin{\left(\frac{1}{x}\right)}.$$

This can be differentiated using the chain rule and the derivative of the inverse sine function. Let

$$u=\frac{1}{x}$$

and find the derivative,

$$\begin{align}\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}u}\arcsin{u}\frac{\mathrm{d}u}{\mathrm{d}x} \\[0.5em] &= \frac{1}{\sqrt{1-u^2}}\frac{\mathrm{d}u}{\mathrm{d}x}. \end{align}$$

Substitute back \(u \) and its derivative to get

$$\frac{\mathrm{d}}{\mathrm{d}x} \frac{1}{x}=-\frac{1}{x^2}.$$

Then work the resulting expression with a bit of algebra to find

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1-\left(\frac{1}{x}\right)^2}}\cdot\left(-\frac{1}{x^2}\right).$$

You can rewrite this last equation by working the expression inside the root and using the fact that the square root of \( x\) squared is equal to the absolute value of \( x\), that is

$$\sqrt{x^2}=|x|.$$

From here you can further simplify the equation to obtain

$$\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{1}{|x|\sqrt{x^2-1}},$$

giving you the derivative of the inverse cosecant function

$$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arccsc}{\, x} =-\frac{1}{|x|\sqrt{x^2-1}}.$$

The derivative of the inverse secant can be found similarly, you just need to use the derivative of the inverse cosine instead.

Graphs of the derivatives of the inverse trigonometric functions

You might have noticed that, unlike the derivatives of trigonometric functions, the derivatives of the inverse trigonometric functions are rational functions that sometimes involve square roots as well. This sure sounds a little extravagant, but the graphs look really cool! Let's take a look at them!

Inverse sine and cosine

When taking a look at the graphs of the derivatives of the inverse trigonometric functions you should pay special attention to their domain. In the case of the inverse sine and inverse cosine, the domain is

$$-1 \leq x \leq 1,$$

so the graph of the derivative of the inverse sine will be shown on the same interval.

Fig. 3. Graph of the derivative of the inverse sine function.

Since the derivative of the inverse cosine is the negative of the graph above, the inverse cosine graph is the inverse sine graph reflected across the x-axis.

Fig. 4. Graph of the derivative of the inverse cosine function.

Note that there are asymptotes at \( x=-1 \) and \( x=1.\)

Inverse tangent and cotangent

This time begin by recalling that the domain of the tangent and cotangent functions are all real numbers, so their graphs extend to infinity. The graph of the derivative of the inverse tangent is given below.

Fig. 5. Graph of the derivative of the inverse tangent function.

Again, the derivative of the inverse cotangent has the opposite sign as the derivative of the inverse tangent, so another reflection across the x-axis is present.

Fig. 6. Graph of the derivative of the inverse cotangent function.

In this case there are no vertical asymptotes!

Inverse secant and cosecant

For the inverse secant and inverse cosecant its is worth noting that the domain has a discontinuity, that is

$$-\infty < x \leq -1 \, \mbox{ and } \, 1 \leq x < \infty,$$

so the graph of their derivative will have a gap for \( -1 < x < 1.\)

Fig. 7. Graph of the derivative of the inverse secant function.

Finally, the graph of the derivative of the inverse cosecant is also a reflection of the derivative of the inverse secant across the x-axis.

Fig. 8. Graph of the derivative of the inverse cosecant function.