`f(x) = (4 - 3x - x^2)/(x^2 - 1)` Find the derivative of the algebraic function.

Textbook Question

Chapter 2, 2.3 - Problem 25 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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

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

You need to use the quotient rule to differentiate `((x-1)/(x+1)):`

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

`f'(x)= (-3x^2 + 3 -2x^3 + 2x -8x+ 6x^2+ 2x^3)/((x^2-1)^2)`

Reducing like terms yields:

`f'(x)= (3x^2 - 6x + 3)/((x^2-1)^2)`

You need to factor out (3):

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

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

Simplify by `(x-1)^2` :

`f'(x)= 3/((x+1)^2)`

Hence, evaluating the derivative of the function, yields `f'(x)= 3/((x+1)^2).`

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