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.