# Differentiating Inverse Trig Functions

This article today will be about differentiating inverse trigonometric functions. They look weird and may take a bit to wrap your head around. Be patient and you will see results in learning these.

### Introduction

The derivatives of arc sin x, arc cos x, and arc tan x can all be memorized. This will help you and problems will be completed faster if you memorize the relationships. Let’s get to some problems so we can see how this works in practice.

### Problem 1

Find the derivative of y = arc sin 4x.

Lets start with a formula. This is the definition of the inverse derivative function.

$$\frac{d}{dx} arc sin~u = \frac{1}{\sqrt{1-u^2}} * \frac{du}{dx}$$

It is a crazy definition but there it is. Let’s substitute the figures from our problem to work on a solution.

In our problem u=4x * 4x and the derivative of 4x = 4.

$$\frac{dy}{dx} = \frac{1}{\sqrt{1-16x^2} * 4}$$

This will of course be:

$$\frac{dy}{dx} = \frac{4}{\sqrt{1-16x^2}}$$

### Problem 2

Find the derivative of y = arc sin x^2.

Once again we apply the same formula as before.

$$\frac{dy}{dx} = arc sin~u = \frac{1}{\sqrt{1-u^2}} * \frac{du}{dx}$$

We substitute the values into our formula just like we did before.

$$\frac{dy}{dx} = \frac{1}{\sqrt{(1-x^2)^2}}(2x)$$

Once again we multiple those expressions together.

$$\frac{dy}{dx} = \frac{2x}{\sqrt{1-x^4}}$$

### Problem 3

Find the derivative of y= arc tan ax^2.

In this expression we can see that (a) is a constant. So treat it like it was some number.

We apply our formula to this.

$$\frac{dy}{dx} = \frac{1}{\sqrt{1+u^2}} * \frac{du}{dx}$$

First thing you should notice is that our formula is different. This is the definition of the inverse tangent function. Once you got that, lets substitute in our values to see what it looks like.

$$\frac{dy}{dx} = \frac{d}{dx}*(ax^2)(1+((ax^2)^2))$$

Keep everything organized well and this is no problem. Now we simplify everything.

$$\frac{2ax}{1+(a^2~x^4)}$$

This is the answer and it was not as bad as it looked. Just remember to substitute everything in the formula correctly and it won’t be an issue.

### Problem 4

What is the slope of y= 4x arc tan 2x at the point x=1/2.

If you want to know the slope of a function then you must find the derivative first. This problem is a little more involved as you can see. We work on one part at a time to keep things straight. The first thing to notice is that this is a combination of two functions. They are 4x and arc tan 2x. Therefore, we apply the old product rule I covered in an earlier lesson. Let’s set that up to see what it looks like.

$$\frac{dy}{dx} = \frac{1}{1+((2x)^2)}(4x) * \frac{d}{dx} * (2x) + arc tan 2x * \frac{d}{dx} * (4x)$$

That’s a mouthful for sure if you tried to say that out loud. None of it is difficult if you stay organized and keep everything where it is supposed to be. Now we simplify.

$$\frac{dy}{dx} = \frac{8x}{1+(4x^2)} + 4 arc tan 2x$$

When simplified it doesn’t look so bad right? Don’t forget about second half of problem. We have to find a value at our point that was given. The point is x=1/2. Substitute that value in.

$$\frac{dy}{dx} = \frac{4}{2} + 4 arc tan~1$$

Simplify this a bit further. Just so it is clear, I try not to take many shortcuts. I want everyone to see how I got to where I did easily.

$$\frac{dy}{dx} = 2 + 4(\frac{\pi}{4})$$

It should be easy enough for anyone now to use a calculator to multiple that out for the final answer.

### Problem 5

Find the derivative of the expression: y= arc cos (1-2x)

To find the derivative of a problem like this we use the definition of the inverse cosine function.

$$\frac{d}{dx} cos^-1~u = \frac{-1}{\sqrt{1-u^2}} \frac{du}{dx}$$

We now have:

$$\frac{dy}{dx} = \frac{-1}{\sqrt{1-(1-2X)^2}} * (-2)$$

This gives us our final form of:

$$+- \frac{2}{\sqrt{x-x^2}}$$

That is a short introduction of the derivatives of inverse trigonometric functions. It will be key for you to memorize the definitions. This will greatly speed up your work. Besides, your professor probably won’t give you the definitions on a test. So you will need to do lots of problems in order to know these definitions by heart.