Question

differentiate \tan ^{-1}\left( \frac{ \sin2x - \cos2x }{ \sin2x +\cos2x }\right) with relation to x

Answer 1

\[\tan ^{-1}\left( \frac{ \sin2x - \cos2x }{ \sin2x +\cos2x }\right)\]

Answer 2

y = (the whole arctan thingy)

Answer 3

would it not be whatever (arctan)' * the fraction thing ?
if let the fraction thing = u
\[\tan^-1(u) = \frac{ 1 }{ 1+u^2 }\]

Answer 4

\[(\tan^{-1} u)' \] my bad

Answer 5

I'm thinking it's either that or it involves the chain rule?..

Answer 6

if you use the chain rule, you can get the answer but the steps are really ugly.... i dont think the examiner would accept the easy way.... there was some method which used the tan 2x identity

Answer 7

Hmmmm I can't think of what identity that would be..

Answer 8

me niether..

Answer 9

i have solved it using the example zepdrix showed me. it works but isnt there a clean and easy way..?

i remember there was

Answer 10

anyway, thnx guys

Answer 11

Oh oh oh I think I got it.

Answer 12

Hold on lemme check my steps really quick.

Answer 13

geeez i can´t wait!!

Answer 14

Omg............ stupid browser just erased all of my text :(

Answer 15

omggggggggggggggggggggggggggggggggggggggggggg

Answer 16

noooooooooooooooooooooooOOOOOOO!

Answer 17

I made a mistake somewhere anyway...
I thought it simplified to tan4x,
but I made a lil error,
I think it simplifies to sec4x-tan4x.

I'm not sure if that will work for us.. hmm

Answer 18

did you divide the num and denom by cos2x first?

Answer 19

I multiplied the numerator and denominator by the conjugate of the denominator.
Then expanded out the square in the numerator,
and applied the Sine and Cosine Double Angle Identities.
Grrr I don't wanna write it out again +_+

Divide by cos2x? is that a good approach maybe? Hmm

Answer 20

i tried using the conjugate too but i wasn´t getting anywhere.. i´ll try once more

Answer 21

\[\Large\bf\sf =\frac{\tan2x-1}{\tan2x+1}\]So dividing by cos2x gives us this or something? hmm

Answer 22

yepp

Answer 23

ok, i got tan4x - sec4x

Answer 24

I have an idea, I'm gonna try using the tangent sum/difference identity.
\[\Large\bf\sf \tan\left(\alpha-\beta\right)\quad=\quad \frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}\]

Answer 25

dude this is not tan this is ARCtan, so r u sure you can use that?

Answer 26

\[\large\bf\sf \tan\left(2x-\frac{\pi}{4}\right)\quad=\quad \frac{\tan2x-\tan\frac{\pi}{4}}{1+\tan2x \tan\frac{\pi}{4}}\quad=\quad \frac{\tan2x-1}{\tan2x+1}\]Ooo I think that works, doesn't it?

Answer 27

This is just to deal with the inside part still.

Answer 28

yep!! i think that´s it woooT!

Answer 29

cool!! thanks a lot zep!

Answer 30

\[\Large\bf\sf \arctan\left(\tan(2x-\pi/4\right)\quad=\quad 2x-\pi/4\]Something like that? :D
yayyy team lol

Answer 31

i see, the arctan and tan cancel out and then you differentiate 2x - pi/4

Answer 32

now this is what i call differentiating with swag!

Answer 33

Ooo nice we got the correct answer.
I checked it in Wolfram just now using the long messy one.
Derivative = 2