f(x) = (x-3)/(x+1)     find f'(0) f(x) = (x-3)/(x+1)     find f'(0)

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You need to find derivative of function using the quotient rule such that:

You need to substitute 0 for x in f'(x) such that:

Hence, evaluating the derivative of the function at yields

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f(x) = (x-3)/(x+1)

Let f(x) = u/v  such that:\

u= x-3  ==>   u' = 1

v= x+1  ==> v'= 1

f'(x) = (u'v- uv')/v^2

       = [1*(x+1) - (x-3)*1]/ (x+1)^2

       = (x+1 - x + 3)/ (x+1)^2

       = 4/(x+1)^2

f'(0) = 4/(0+1)^2 = 4/1 = 4

==> f'(0) = 4

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