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How to derive the derivative formula of arctan x?

  • Thread starter dnt
  • Start date
  • #1
dnt
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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 im 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.
 

Answers and Replies

  • #2
1,085
6
Do you know the Inverse function theorem? Using it and a right triangle (as you said), gives the result pretty quickly.
 
  • #3
478
2
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?
 
  • #4
dnt
238
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the opposite would be x, the adjacent is 1 and the hypoteneuse is the square root of 1+x^2

after that im stuck.
 
  • #5
Galileo
Science Advisor
Homework Helper
1,989
6
You got tan y = x, you want dy/dx, so use implicit differentiation.
 
  • #6
Gib Z
Homework Helper
3,346
5
[tex]y=arctan x[/tex]

Therefore [tex]x=\tan y[/tex], Quite easy to see.

[tex]\frac{dx}{dy}=\sec^2 y[/tex]

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

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

[tex]\frac{dy}{dx}=\frac{1}{x^2+1}[/tex]
 
  • #7
dextercioby
Science Advisor
Homework Helper
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[tex] y=\arctan x [/tex]

[tex] \tan y =x [/tex]

[tex] \frac{d}{dx}\tan y= \frac{1}{\cos^{2}y}\frac{dy}{dx}=1 [/tex]

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

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