Question

Hi all! How do I take the derivative of the inverse tangent of (x^2) ?

Answer 1

Do you know the derivative of the inverse tangent of x?

Answer 2

yes

Answer 3

1/(1+x^2)

Answer 4

so is it 1/(1+x^4) ?

Answer 5

yes, almost but remeber that when taking the derivative of example sin(x^2) it isn't cos(x^2) but you also have to multiply with the derivative of x^2 so it's 2x*cos(x^2). Can you apply that to the arctan case?

Answer 6

hmm

Answer 7

are you talking about the chain rule?

Answer 8

Precisely, when we take the derivative of f(x^2) we say x^2=u => f(x^2) = f(u) and then the derivative is D(u)*D(f(u))

Answer 9

For the arctan case you have to multiply what you got with the derivative of x^2 = 2x

Answer 10

ah, ok

Answer 11

hmm so it is 1/(1+2x^2 *2x)

Answer 12

ok, so generally we have as above according to the chain rule: D(u)*D(f(u)) and in this case f(u) = arctan(u). The derivative of arctan(u) is as you stated above = 1/(u^2+1) and the derivative of u = 2x. Which means we have D(u)*D(f(u)) = 2x*(1/((x^2)^2+1)) = 2x/(1/((x^4)+1))

Answer 13

sorry I ment 2x/((x^4)+1))

Answer 14

hmm

Answer 15

I thought I would just take the chain rule of the bottom

Answer 16

where (x^2)^2

Answer 17

I see, it makes sense

Answer 18

obviously I'm dealing with the entire function

Answer 19

Yes, precisely

Answer 20

cool, thanks!

Answer 21

No problem :)