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Derivatives of Inverse Trigonometric Functions

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What would you do if you need to fix something? This question is rather general, but depending on the scenario you will need an appropriate **tool **(or tool set)** **to do the job. Something similar happens in mathematics. There are lots of tools that can be used to our convenience. A particularly nice set of tools are the** Inverse Trigonometric Functions**!

Asking for the derivative of inverse trigonometric functions is a common task in **differential calculus**, but it also plays a major role in **integral calculus** where you use the inverse trigonometric functions as tools for finding some integrals. For this reason, let's look at how to find the derivatives 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.

The **inverse sine **function is also known as the **arcsine** function. There are two equivalent notations for this function:

$$\sin^{-1}{x}\equiv\arcsin{x}.$$

The rest of the inverse trigonometric functions are denoted similarly:

$$\cos^{-1}{x}\equiv\arccos{x},$$

$$\tan^{-1}{x}\equiv\arctan{x},$$

$$\cot^{-1}{x}\equiv\mathrm{arccot}{\,x},$$

$$\sec^{-1}{x}\equiv\mathrm{arcsec}{\,x},$$

and

$$\csc^{-1}{x}\equiv\mathrm{arccsc}{\,x}.$$

Remember that \( \equiv \) means that the two things are equivalent. In other words they are exactly the same thing.

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.

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

The derivatives of the inverse trigonometric functions are given as follows:

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

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

$$\frac{\mathrm{d}}{\mathrm{d}x}\arctan{x}=\frac{1}{1+x^2},$$

$$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arccot}{\,x}=-\frac{1}{1+x^2},$$

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

and

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

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.

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

Use the above differentiation rule(s).

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!

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!

Find the derivative of \( f(x)=\arcsin{x^2}.\)

Answer:

*Identify which differentiation rule is relevant.*

The function is written as a composition of functions and there are no products or quotients involved, so you can do this derivative using **the chain rule.**

2. *Use the differentiation rule, which in this case is the **chain rule.*

Since you are using the chain rule, you should start by letting \(u=x^2\) and then apply the chain rule, so

$$f'(x)=\left( \frac{\mathrm{d}}{\mathrm{d}u}\arcsin{u} \right)\cdot \frac{\mathrm{d}u}{\mathrm{d}x}.$$

3. *W**rite the derivatives of the functions involved in the calculation.*

You can now write the derivative of the inverse sine function in the above expression

$$f'(x)=\frac{1}{\sqrt{1-u^2}}\cdot \frac{\mathrm{d}u}{\mathrm{d}x}.$$

You will also need to find the remaining derivative. Since \(u=x^2,\) you can find its derivative using the power rule,

$$\frac{\mathrm{d}u}{\mathrm{d}x}=2x,$$

and then substitute it back, so

$$f'(x)=\frac{1}{\sqrt{1-u^2}}\cdot 2x.$$

Whenever you make a change of variable, you need to undo it at the end, so substitute back \( u=x^2 \) and simplify, that is

$$\begin{align}f'(x) &= \frac{1}{\sqrt{1-\left( x^2 \right)^2}}\cdot 2x \\[0.5em] f'(x) &= \frac{2x}{\sqrt{1-x^4}}.\end{align}$$

How about the product rule?

Find the derivative of \(g(x)=\left(\arctan{x}\right) \left(\cos{x}\right). \)

Answer:

1. *Identify which differentiation rule is relevant.*

The function is written as a product of functions, hence you need to use **the product rule**.

2. *Use the differentiation rule, in this case the product rule*

The products involved are the inverse tangent function and the cosine function, so

$$g'(x)= \left( \frac{\mathrm{d}}{\mathrm{d}x}\arctan{x} \right) \cos{x} + \arctan{x} \left( \frac{\mathrm{d}}{\mathrm{d}x}\cos{x} \right).$$

3. *Write the derivatives of the functions involved in the calculation.*

You can find above the derivative of the inverse tangent function, and the derivative of the cosine function is the negative of the sine function, so

$$\begin{align}g'(x) &= \left( \frac{1}{1+x^2} \right)\cos{x} + \arctan{x} \left( -\sin{x} \right) \\[0.5em] &= \frac{\cos{x}}{1+x^2}-\left(\arctan{x}\right) \left(\sin{x} \right). \end{align}$$

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.

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

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

Let's go back into the proof of the derivative of the inverse sine function. After doing the implicit differentiation you were left with the following equation:

$$\cos{y}\frac{\mathrm{d}y}{\mathrm{d}x}=1.$$

If you substitute back \( y=\arcsin{x} \) you will have a composition of a trigonometric function and an inverse trigonometric function, that is

$$\cos{\left(\arcsin{x}\right)}.$$

There is a neat method where you can use an auxiliary triangle to find this composition. First, build a triangle using \(\sin{y}=x,\) which means that the ratio of the opposite leg to the hypotenuse is equal to \(x.\) This idea is better understood if you write it as

$$\begin{align} \sin{y} &= x\\[0.5em] &= \frac{x}{1}.\end{align}$$

Here you have to look at \( y \) as if it was an angle.

The remaining leg can be found by using the Pythagorean Theorem

$$a^2+b^2=c^2,$$

where \(a=x,\) \(c=1,\) and \( b \) is the missing leg, so

$$\begin{align} b &= \sqrt{c^2-a^2} \\ &= \sqrt{1-x^2}. \end{align}$$

Now that you know the length of the adjacent leg, you can write the cosine of \(y\) as the ratio of the adjacent leg and the hypothenuse.

$$\begin{align} \cos{y} &= \frac{\sqrt{1-x^2}}{1} \\ &= \sqrt{1-x^2}.\end{align}$$

With this information you can now write the derivative of the inverse sine function,

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

Try doing this with the derivatives of the other inverse trigonometric functions!

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

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.

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!

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.

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.

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

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.

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.

In this case there are no vertical asymptotes!

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 \( -1 < x < 1.\)

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.

- The inverse of the sine function is known as the arcsine function. The rest of the inverse trigonometric functions are named in a similar way.
- The derivatives of the six inverse trigonometric functions are the following:
- $$\frac{\mathrm{d}}{\mathrm{d}x}\arcsin{x}=\frac{1}{\sqrt{1-x^2}}.$$
- $$\frac{\mathrm{d}}{\mathrm{d}x}\arccos{x}=-\frac{1}{\sqrt{1-x^2}}.$$
- $$\frac{\mathrm{d}}{\mathrm{d}x}\arctan{x}=\frac{1}{1+x^2}.$$
- $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arccot}{\,x}=-\frac{1}{1+x^2}.$$
- $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arcsec}{\,x}=\frac{1}{|x|\sqrt{x^2-1}}.$$
- $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arccsc}{\,x}=-\frac{1}{|x|\sqrt{x^2-1}}.$$

- The derivatives of the inverse trigonometric functions can be proved by using implicit differentiation and applying Pythagorean trigonometric identities.
- An auxiliary triangle can be used if you are struggling to remember the Pythagorean trigonometric identities.

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