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

Question

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

hpfan101 · Answer

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

hpfan101 · Answer

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

freckles · Answer

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

hpfan101 · Answer

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

freckles · Answer

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

freckles · Answer

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

hpfan101 · Answer

Ok, will do.

hpfan101 · Answer

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

freckles · Answer

yep looks great!

freckles · Answer

a=b=c=1

hpfan101 · Answer

Cool, thanks for the help!