The derivative of the arctan of x is equal to 1 divided by (1+x^{2}).

Hey, guys, I’m gonna show you. How to find the derivative arctan/tan-inverse of X.

y = tan−1 x

tan y = tan(tan−1 x)

tan y = x

So, we’re gonna

Let, the derivative of y = arctan of x

y’ = 1/(1+x^{2})

and now we apply the function of tan to both sides to get tan y equals

tan of arctan of X but tan and arctan

inverse functions. So, they cancel out so,

we get tan of y equals x and now we want

to implicitly differentiate so the

derivative of tan is sec2 this is

y. and then we use a chain rule so the

derivative of Y is dy/dx and this equals 1 So, let’s rearrange for dy/dx this

equals 1 over second square. I’m just

gonna write something here do the

derivative equals x squared of Y. But we want to find the derivative of arctan of x in terms of X’s at the moment we’ve got away but here we’ve got a relationship that expresses Y’s in terms of X’s so we want to express sec

of Y in terms of tan away and then we

can substitute in for X so we know that

sine squared of y plus cosine is going

to y equals 1 but if we divide every

single term by cos squared away we’re

gonna get tan squared y plus 1 equals

SEC squared of Y and this is another

identity so now we can substitute in for

sec^{2} so this is gonna equal 1 over tan^{2} of y+1 and we know

that tan of y= x so the last step

is substitute in Y so this gives us 1 over x+1 so there you go the derivative of arctan of X.