How do you differentiate $f(x)= (x-1)/(x+1)$ ?

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Answer 1 · 2016-05-25T13:53:37

$f' (x)=2/(x+1)^2$

Use $d/dx(u/v)=(v*d/dxu-u*d/dx(v))/(v^2)$

Let $u=x-1$ and $v=x+1$

$d/dx(u/v)=(v*d/dxu-u*d/dx(v))/(v^2)$

$f' (x)=d/dx((x-1)/(x+1))$

$f' (x)=((x+1)*d/dx(x-1)-(x-1)*d/dx(x+1))/((x+1)^2)$

$f' (x)=((x+1)(1)-(x-1)(1))/((x+1)^2)$

$f' (x)=(x+1-x+1)/(x+1)^2$

$f' (x)=2/(x+1)^2$

God bless....I hope the explanation is useful.

How do you differentiate f(x)= (x-1)/(x+1)f(x)= (x-1)/(x+1)?