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.

MR
Answered by Matthew R. Maths tutor

12809 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Why does d/dx (tan(x)) = sec^2(x)?


Differentiate y=x^x with respect to x.


Express 4sinx + 3cosx in the form Rcos(x-a)


Expand and simplify (3 + 4*root5)(3 - 2*root5)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences