# 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 **

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**

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

by putting 0 on the value of x we get,

f(0) = (0-3)/(0+1)

f(0) =-3/1

f(0) =-3

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

As, x= 0

(plugging the values in the equation)

f(0) = (0-3) / (0+1)

f(0) = (-3) / (+1)

f(0) = -3

To calculate f'(0), we'll use the definition of the derivative:

f'(0) = lim {[f(x) - f(0)]/(x-0)} , x-->0

f'(0) = lim {[(x-3)/(x+1) - (-3)/1]/ x}, x-->0

f'(0) = lim {[(x-3)/(x+1) - (-3)/1]/ x}, x-->0

f'(0) = lim [(x-3 + 3x + 3)/ x(x+1)], x-->0

We'll combine like terms:

f'(0) = lim [(x-3 + 3x + 3)/ x(x+1)], x-->0

f'(0) = lim [4x/x(x+1)], x-->0

f'(0) = lim [4x/x(x+1)], x-->0

f'(0) = lim [4/(x+1)], x-->0

f'(0) = lim [4/(x+1)], x-->0

**f'(0) = 4/(0+1) = 4/1 = 4**

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

To find f'(0).

Solution:

f(x) = (x+3)/(x+1) = [(x+1)- 4]/(x-1)

f(x) = (x+1)/(x+1) - 4/(x+1).

f(x) = 1 - 4/(x+1)

f'(x) = (1)' - {4/(x+1)}'

f'(x) = 0 - 4/(x+1)^2, as (constant)' = 0 and (k/(x+a)^n)' = - kn/(x-a)^(n+1).

Therefore, f'(0) = -4/(0+1)^2 = -4.

Given:

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

Let:

x - 3 = u

And

x + 1 = v

Then:

f(x) = u/v

And:

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

u' = 1

v' = 1

Substituting values of u, v, u' and v' in equation for f"(x):

f'(x) = [1*(x + 1) - 1*(x - 3)]/[(x + 1)^2]

= (x + 1 - x + 3)/[(x + 1)^2]

= 4/[(x + 1)^2]

Therefore:

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

= 4/1 = 4

Answer:

f'(0) = 4