# f(x) = (x-3)/(x^2+2) find f ' (1) .

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f(x) = (x-3)/(x^2 + 2)

Let f(x) = u/v such that:

u= x-3 ==> u' = 1

v= x^2 + 2 ==> v' = 2x

We know that:

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

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

= (x^2 + 1 - 2x^2 + 6x)/(x^2 + 2)^2

= (-x^2 + 6x + 1)/(x^2 + 2)^2

Now subsitute with x= 1:

==> f'(1) = (-1+6+1)/(1+2)^2

= 6/9 = 2/3

**==> f'(1) = 2/3**

We'll use the limit method to calculate the value of derivative of a function in a given point.

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

lim [(x-3)/(x^2+2) + 2/3]/(x-1) = lim (3x-9+2x^2+4)/(x-1)

lim (3x-9+2x^2+4)/(x-1) = lim (3x-5+2x^2)/(x-1)

We'll substitute x by 1:

lim (3x-5+2x^2)/(x-1) = (3-5+2)/(1-1) = 0/0

Since we've obtained an indeterminacy, we'll apply L'Hospital rule:

lim (3x-5+2x^2)/(x-1) = lim (3x-5+2x^2)'/(x-1)'

lim (3x-5+2x^2)'/(x-1)' = lim (3+4x)/1

We'll substitute x by 1:

lim (3+4x)/1 = (3+4)/1

f'(1) = lim (3x-5+2x^2)/(x-1) = 7

**f'(1) = 7**

f(x) = (x-3)/(x^2+2)

To find f'(x).

Solution:f(x) is in the form u(x)/v(x).

Therefore f'(x) = {u'(x)v(x) +u(x)v'(x)}/(v(x))^2

u(x) = x-3. So u'(x) = (x-3)' = 1.

v(x) = x^2+2). Therefore v'(x) = (x^2+2)' = 2x.

Therefore f'(x) = {(x-3)'(x^2+2) +(x-3)(x^2+2)}/(x^2+3)^2.

f'(x) = {x^2+1 - (x-3)(2x)}/(x^2+2)^2.

f'(x) = {x^2+2 - 2x^2+6x}/(x^2+3)^2.

f'(x) = (6x-x^2)/(x^2+3).

Now put = 1 to getf'(1):

f'(1) = (6*1-1^2)/(1^2+2)^2

f'(1) = 5/9