Expert: Socrates - 8/24/2006

Question
What is the derivative in respect to x of

tan^(-1)x

I don't believe this is very complicated.  It's just been about a year since I've had an calculus and I am at a loss as to how to approach this.  Thank you.

Answer
The answer is 1/(1+x^2)

You can't use the rules for taking derivatives to get this directly, you must derive it by a special argument. For most students, it is best just to memorize the derivative.

To see why it is 1/(1+x^2), use the chain rule to take the derivative of both sides of the equation

tan(tan^-1(x))) = x

sec^2(tan^-1(x)) (tan^-1(x)))' = 1

if the tangent of an angle is x , you can draw a right triangle with legs having length x and 1 and hypotenuse (1+x^2)^1/2 . This shows that sec(tan^-1(x)))=(1+x^2)^1/2

Then sec^2(tan^-1(x)) = 1 + x^2

so we now have (1+x^2) (tan^-1(x)))' = 1

divide both sides by (1+x^2) and get

(tan^-1(x)))' = 1/(1+x^2)

which is the correct answer