How to derive the derivative formula of arctan x?

[dnt]

Homework Statement

basically what the topic states - derive the formula for the derivative of arctan x.

Homework Equations

d/dx (arctan x) = 1/(1+x^2)

The Attempt at a Solution

strange question because we already know the answer. but I am not sure how to start this.

i know arctan x = y

therefore tan y = x

but what can i do with this? do i need to draw a right triangle and label all the sides? can someone help me get started? thanks.

 

[mattmns]

Do you know the Inverse function theorem? Using it and a right triangle (as you said), gives the result pretty quickly.

 

[z-component]

You could indeed draw a triangle knowing that the tangent is equal to x. In a triangle, if you're given that the tangent of some angle y is equal to x, what do you know about the lengths of the sides?

 

[dnt]

the opposite would be x, the adjacent is 1 and the hypoteneuse is the square root of 1+x^2

after that I am stuck.

 

[Galileo]

You got tan y = x, you want dy/dx, so use implicit differentiation.

 

[Gib Z]

y=arctan x

Therefore x=\tan y, Quite easy to see.

\frac{dx}{dy}=\sec^2 y

Using the Pythagorean Identity \sec^2 y = \tan^2 y +1 we can get this: \frac{dx}{dy}=\tan^2 y +1.

Flip the fraction since we want dy/dx, not dx/dy. And also, as seen on my second line tan y=x, so tan^2 y = x^2.

Thats how we get

\frac{dy}{dx}=\frac{1}{x^2+1}

 

[dextercioby]

y=\arctan x

\tan y =x

\frac{d}{dx}\tan y= \frac{1}{\cos^{2}y}\frac{dy}{dx}=1

\frac{dy}{dx} =\cos^{2} y=\left(\frac{1}{\sqrt{1+x^{2}}}\right)^{2}=\frac{1}{1+x^{2}}