Inverse trig (1 Viewer)

[nahi11]

Hi,

Could anyone please this inverse trig with working out.

tan^-1 (1/1-x) [So essential inverse tan 1/1-x]

Thanks.

Edit: Sorry the question is "Find the derivative of the above function in simplest form"

 

[Timske]

I'm not sure what the question is

 

[SpiralFlex]

What do you want? Simplification?

 

[nahi11]

Sorry, the question was to find the derivative.

 

[SpiralFlex]

Edit: Blergh.

 

[SpiralFlex]

Wait sorry I am wrong. OMG

 

[RealiseNothing]

Quoting for evidence that spiral has been wrong in his life atleast once.

 

[nahi11]

The answer is 1/ x^2 -2x +x

which is also

1/(x-1)^2 + 1

Also, Spiral what software/program do you use to display your working out like that?

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{d}{dx} \tan^{-1}\left (\frac{1}{1-x} \right ) = \frac{1}{x^2 - 2x @plus; 2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{d}{dx} \tan^{-1}\left (\frac{1}{1-x} \right ) = \frac{1}{x^2 - 2x + 2}" title="\frac{d}{dx} \tan^{-1}\left (\frac{1}{1-x} \right ) = \frac{1}{x^2 - 2x + 2}" /></a>

 

[nahi11]

lol thanks, can you please show me the working out? nevermind, saw spirals post. Thanks for the help guys.


The problem was that I used the formula from the Cambraidge 3U textbook. d/dx tan x/a = a/a^2 + x^2

 

[SpiralFlex]

Sorry about my mix up. >_> *Slaps*