How to find the derivative of arctan(x)

Let y = arctan(x). Then x = tan(y).

Differentiate using the chain rule and rearrange: d(x)/dx = d(tany)/dx So 1 = sec^2(y) * dy/dx dy/dx = 1/sec^2(y)

But from identity sin^2(y) + cos^2(y) = 1 We can derive, by diving across by cos^2(y) tan^2(y) + 1 = sec^2(y)

So by combining: dy/dx = 1/sec^2(y) sec^2(y) = tan^2(y) + 1 y = arctan(x)

Get the result: dy/dx = 1/(x^2 + 1)

Maybe attempt to do this for a general inverse function: y = f^-1(x) y' = (f'(f^-1(x)))^-1 With geometric explanation from graph.