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

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### 2 Answers

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

Let f(x)= u/v

==> u= x-2 ==> u'= 1

==> v= x^2 + 3 ==> v'= 2x

Then use the chain rule:

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

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

= (x^2 +3 -2x^2 +4x)/(x^2+3)^2

= (-x^2 +4x +3)/(x^2 + 3)^2

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

Solution:

f(x) = u(x)*v(x), f'(x) = u(x)v'(x) + u'(x)v(x)

Here u(x) = x-2 and v(x) = 1/(x^2+3)

u'(x) = (x-2) =1, v'(x) {1/(x^2+3)^2} = -(1/(x^2+3))(x^2)' = -2x/(x^2+3), {u(v(x))' = u'(x)*v'(x)

So f(x) = (x-2){-2x/(x^2+3)^2 +1/(x^2+3)

= {(x-2)(-2x) +(x^2+3)}/(x^2+3)^2

={-2x^2+4x+x^2+3}/(x^2+3)

= (-x^2+4x+3)/(x^2+3)^2

=-(x^2-4x-3)/(x^2+3)