# Find the derivative of y= (2x-3) / (x^2-1)

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### 2 Answers

We need to find the derivative of y = (2x - 3)/(x^2 - 1)

y = (2x - 3)/(x^2 - 1)

=> y = (2x - 3)(x^2 - 1)^(-1)

y' = 2*(x^2 - 1)^(-1) - (2x)(2x - 3)*(x^2 - 1)^(-2)

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

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

=> [-2x^2 + 6x - 2]/(x^2 - 1)^2

**The required derivative is [-2x^2 + 6x - 2]/(x^2 - 1)^2**

y= (2x-3) / (x^2 -1)

We will use the quotient rule to find the derivative.

Let y= u/ v such that"

u= 2x-3 ==> u' = 2

v = x^2 -1 ==> v' = 2x

Then we know that:

y' = ( u'v- uv' ) / v^2

==> y' = ( 2*(x^2-1) - (2x(2x-3)]/ (x^2-1)^2

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

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

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