## Derivative of Arctan Lesson

### How do you Differentiate Arctangent?

The inverse tangent - known as arctangent or shorthand as arctan, is usually notated as tan^{-1}(*some function*). To differentiate it quickly, we have two options:

**Use the simple derivative rule.****Derive the derivative rule, and then apply the rule.**

In this lesson, we show the derivative rule for tan^{-1}(u) and tan^{-1}(x). Additionally, we cover how the derivative rule is derived.

At the end of the lesson, there are four example problems to help with your understanding of this subject.

#### Derivative of Arctan(u)

The derivative rule for arctan(u) is given as:

$$\begin{align} & [tan^{-1}(u)]' = \frac{u'}{1+u^{2}} \end{align}$$

Where *u* is a function of a single variable, and the prime symbol * '* denotes the derivative with respect to that variable. Here are some examples of a single variable function

*u:*

- u = x
- u = sin(x)
- u = y
^{3}- 3y + 4

#### Derivative of Arctan(x)

The derivative rule for arctan(x) is the arctan(u) rule but with each instance of *u* replaced by *x*. Since the derivative of x is 1, the numerator simplifies to 1. The derivative rule for arctan(x) is given as:

$$\begin{align} & [tan^{-1}(x)]' = \frac{1}{1+x^{2}} \end{align}$$

Where * '* denotes the derivative with respect to x.

### What Makes Arctan Differentiable?

Arctan is a differentiable function because **its derivative exists on every point of its domain**. In the image below, a single period of arctan(x) is shown graphed. The curve is continuous and does not have any sharp corners.

If there is a sharp corner on a graph, the derivative is not defined at that point. So, if you come across a function whose graph has sharp corners, it will not be differentiable on every point of its domain.

#### Proof of the Derivative Rule

Since arctangent means inverse tangent, we know that arctangent is the inverse function of tangent. **Therefore, we may prove the derivative of arctan(x) by relating it as an inverse function of tangent.** Here are the steps for deriving the arctan(x) derivative rule.

- y = arctan(x), so x = tan(y)
^{dx}⁄_{dy}[x = tan(y)] = sec^{2}(y)- Using sum of squares corollary: sec
^{2}(y) = 1 + tan^{2}(y) - tan
^{2}(y) = x^{2}, so^{dx}⁄_{dy}= 1 + x^{2} - Flipping
^{dx}⁄_{dy}, we get^{dy}⁄_{dx}=^{1}⁄_{(1 + x2)}

### Derivative of Arctan Example Problems

#### Derivative of Arctan(2x)

$$\begin{align} & \text{Find the derivative of } tan^{-1}(2x) \text{ with respect to } x \text{.} \\ \\ & \text{Solution:} \\ \\ & \text{1.) Let's set } u = 2x \\ \\ & \text{2.) } u' = 2 \\ \\ & \text{3.) } u^{2} = 4x^{2} \\ \\ & \text{4.) } \frac{d}{dx}\left[tan^{-1}(2x)\right] = \frac{2}{1+4x^{2}} \\ \\ & \Rightarrow \text{The solution is } \frac{2}{1+4x^{2}} \end{align}$$

#### Derivative of Arctan(^{1}⁄_{x})

$$\begin{align} & \text{Find the derivative of } tan^{-1}\left(\frac{1}{x}\right) \text{ with respect to } x \text{.} \\ \\ & \text{Solution:} \\ \\ & \text{1.) Let's set } u = \frac{1}{x} \\ \\ & \text{2.) } u' = -\frac{1}{x^{2}} \\ \\ & \text{3.) } u^{2} = \frac{1}{x^{2}} \\ \\ & \text{4.) } \frac{d}{dx}\left[tan^{-1}\left(\frac{1}{x}\right)\right] = \frac{-\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}} \\ \\ & \text{5.) Simplify: } \frac{-\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}} \cdot \frac{x^{2}}{x^{2}} = -\frac{1}{x^{2}+1} \\ \\ & \Rightarrow \text{The solution is } -\frac{1}{x^{2}+1} \end{align}$$

#### Derivative of Arctan(4x)

$$\begin{align} & \text{Find the derivative of } tan^{-1}\left(4x\right) \text{ with respect to } x \text{.} \\ \\ & \text{Solution:} \\ \\ & \text{1.) Let's set } u = 4x \\ \\ & \text{2.) } u' = 4 \\ \\ & \text{3.) } u^{2} = 16x^{2} \\ \\ & \text{4.) } \frac{d}{dx}\left[tan^{-1}\left(4x\right)\right] = \frac{4}{1+16x^{2}} \\ \\ & \Rightarrow \text{The solution is } \frac{4}{1+16x^{2}} \end{align}$$

#### Derivative of Arctan(x^{2} + 1)

$$\begin{align} & \text{Find the derivative of } tan^{-1}\left(x^{2}+1\right) \text{ with respect to } x \text{.} \\ \\ & \text{Solution:} \\ \\ & \text{1.) Let's set } u = x^{2}+1 \\ \\ & \text{2.) } u' = 2x \\ \\ & \text{3.) } u^{2} = \left(x^{2}+1\right)\left(x^{2}+1\right) = x^{4}+2x^{2}+1 \\ \\ & \text{4.) } \frac{d}{dx}\left[tan^{-1}\left(x^{2}+1\right)\right] = \frac{2x}{1+x^{4}+2x^{2}+1} \\ \\ & \text{5.) Simplify: } \frac{d}{dx}\left[tan^{-1}\left(x^{2}+1\right)\right] = \frac{2x}{x^{4}+2x^{2}+2} \\ \\ & \Rightarrow \text{The solution is } \frac{2x}{x^{4}+2x^{2}+2} \end{align}$$