Question

Differentiate:
\[f(x)=\arctan(\frac{ x-1 }{ x+1 }) \]

Answer 1

\[f'(x)=\frac{ 1 }{ 1+(\frac{ x-1 }{ x+1 })^2 }\times(\frac{ x-1 }{ x+1 })'\]

Answer 2

I'm not sure how to simplify the expression after taking the entire derivative.

Answer 3

for the fraction there you could multiply top and bottom by (x+1)^2
giving you...
\[f'(x)=\frac{(x+1)^2}{(x+1)^2+(x-1)^2} \cdot (\frac{x-1}{x+1})'\]

Answer 4

Oh, ok, thanks! I think I know what to do for the rest of the problem.

Answer 5

your answer should turn into a really pretty answer at then end...

Answer 6

Here is a hint:
it will be of the form:
\[\frac{a}{bx^2+c} \\ \text{ where } a,b,c \text{ are constants }\]
if you do not end up with this form
please show me what you have

Answer 7

Ok, will do.

Answer 8

My final answer is:
\[f'(x)=\frac{ 1 }{ x^2+1 }\]

Answer 9

yep looks great!

Answer 10

a=b=c=1

Answer 11

Cool, thanks for the help!