# Differentiate f(x) = (3x-2)/(x^2+1) and calculate f'(1)

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### 1 Answer

We have to differentiate f(x) = (3x-2)/(x^2+1) and find f'(1).

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

Using the chain and the product rules.

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

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

=> (-6x^2 - 4x + 3x^2 + 3)/(x^2 +1 )^2

=> (-3x^2 - 4x + 3)/ (x^2 +1 )^2

**f'(x) = (-3x^2 + 4x + 3)/ (x^2 +1 )^2**

So f'(1) = (-3*1^2 + 4*1 + 3)/ (1^2 +1 )^2

=>(-3 + 4 + 3)/ 4

=> 4/4

=> 4/4

**The required result is 1.**