Derivative of arctan(x) is:

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

Let’s see the proof!

We can start with arctan(x):

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\[y=tan^-^1(x)\]

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

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\[tan y=tan [tan^-^1(x)] \]

and then it follows that:

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\[tany=x\] if \[\frac{-\pi}{2} < y < \frac{\pi}{2}\]

We can now use implicit differentiation to find y’ in terms of y. Remember that y=arctan(x), so our end goal is to find y’ in terms of x.

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

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

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

So, y’=cos^{2}(y).

Now we need to find the derivative in terms of x. To do so, we find cos(y) in terms of x and plug it into y’=cos^{2}(y). We can use our original relationship y=arctan(x) to draw a triangle, and use the pythagorean theorem to find the hypotenuse. In the diagram below, y is the angle in the top left corner.

This triangle gives us the following relationships:

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

Since we have cos(y) in terms of x, we can square cos(y) and plug it into y’:

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

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

Let’s 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:

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

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

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

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