# Deriving formula of arctan x derivative

• Dec 11th 2009, 06:51 PM
ughintegrals
Deriving formula of arctan x derivative
hi all.
im kinda struggling with this assignment question.

Starting from tan(arctan x) = x, derive the formula (arctan x)' = 1/(1+x^2)

i understand how to get to the formula from a simple arctan x = x, but not with that tan in front.

any suggestions?
• Dec 11th 2009, 06:57 PM
Jameson
Quote:

Originally Posted by ughintegrals
hi all.
im kinda struggling with this assignment question.

Starting from tan(arctan x) = x, derive the formula (arctan x)' = 1/(1+x^2)

i understand how to get to the formula from a simple arctan x = x, but not with that tan in front.

any suggestions?

arctan(x) is an angle, right? So if the tangent of that angle is x, you can construct a right triangle to demonstrate this, with the ratio x:1. You can also solve for the hypotenuse now.

Take the derivative of the entire equation.

$\sec^2 \Big[ (\arctan(x)) \Big] *(\arctan(x))'=1$

Solve for arctan(x) prime and remember that arctan(x) is an angle.
• Dec 11th 2009, 07:24 PM
ughintegrals
i hate to be difficult, but i dont entirely understand.
is there anyway you could elaborate?

im particularly stuck on how that equation was formulated...
thanks again.
• Dec 11th 2009, 07:33 PM
Jameson
Quote:

Originally Posted by ughintegrals
i hate to be difficult, but i dont entirely understand.
is there anyway you could elaborate?

im particularly stuck on how that equation was formulated...
thanks again.

It's ok. I'll try to explain more.

The tangent of an angle is the ratio of the opposite side to the adjacent side. You know this from geometry surely. Inverse trig functions, like arctan(x) take those ratios and tell you what angle has that ratio, thus the result of arctan(x) will be an angle.

For simplicity, say arctan(x)=y. The you start with tan(y)=x. Draw a right triangle in the first quadrant with the angle between the x axis and the hypotenuse being y. Now if the tangent of this angle, y, is equal to x, that means that the ratio of the opposite leg to the adjacent one is x, or x/1. So the y-leg of the right triangle is x and the horizontal leg is 1.

Now use the Pythagorean Theorem to find the hypotenuse. Now for the derivative.

What is [tan(y)]' ? It's sec^2(y)*y' by the chain rule. And the derivative of x is just 1. That's where I got my equation from. Solve for y' to get y' = 1 / (sec^2(y)).

Remember sec x = 1/(cos(x)) so find the cosine of the angle, flip that to get the secant, then square it and finish the problem.
• Dec 11th 2009, 08:28 PM
ughintegrals
*clicks!*

I finally got it! It just took me a bit to worrk it all out and understand..

Now, looking back, its not so difficult!(Surprised)(Giggle)

Thank you VERY very much for your time, help and consideration.
It truly means a lot.

Cheers, and God bless!