Derivative of arctan(x) is:

    \[\frac{d}{dx}arctanx=\frac{1}{1+x_2}\]

Let’s start with tanx:

    \[y=tan^-^1(x)\]

We can simplify the equation by taking the tangent of both sides:

    \[tan y=tan [tan^-^1(x)] \]

and then it follows that:

    \[tany=x\]

and

    \[\frac{-\pi}{2} < y < \frac{\pi}{2}\]


Notice that if y=arctan(x), then:

    \[\sin=\frac{x}{\sqrt{1+x^2}},  cosy=\frac{1}{\sqrt{1+x^2}}\]

y=arctan(x) derivative of arctan(x) right triangle 1 x

By using Implicit Differentiation:

    \[ \frac{d}{dx}[tan(y)]=\frac{d}{dx}x\]

    \[ sec^2y.\frac{dy}{dx}=1\]

    \[ \frac{dy}{dx}=\frac{1}{sec^2y}=cos^2y\]

In other words, y’=cosy.

Yet, actually, we aim to find the derivative as a function of x, not of y. Let’s use the equation y=tan-1(x). We get y’=cos2(arctan(x)). We can simplify it by using some geometry theorems.

geometry thorem right angle  Pythagorean theorem


In this triangle, tan(y)=x , so y=arctan(x). And then, 

    \[ cos(y)=\frac{1}{\sqrt{1+x^2}}\]

    \[cos^2(y)=\frac{1}{1+x^2} \Rightarrow \frac{dy}{dx}=\frac{1}{1+x^2}\]

In other words, we can conclude that:

    \[ \frac{d}{dx}arctan(x)=\frac{1}{1+x^2}\]

Lets continue with an example. Below, there is an equation with an arc in it. We will try to take derivative of this expression. We will use of both the basic rules of the derivative and the arctan derivative we mentioned above.

    \[y=arctan\frac{x+1}{x-1}, x\neq 1\]

Now, here is the step by step solution:

    \[y'(x)=(arctan\frac{x+1}{x-1})'=\frac{1}{1+(\frac{x+1}{x-1})^2}.(\frac{x+1}{x-1})' \]

    \[ =\frac{1.(x-1)-(x+1).1}{(x-1)^2 + (x+1)^2}\]

    \[ =\frac{x-1-x-1}{x^2-2x+1+x^2+2x+1}\]

    \[ \frac{-2}{2x^2+2} = \frac{-1}{1+x^2}\]