You need to find derivative of the given function using the quotient rule and the chain rule such that:

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

`f'(x) = (0*(x^2-3x)^2- 4*2(x^2-3x)(2x - 3))/((x^2-3x)^4)`

`f'(x) = (-8(x^2-3x)(2x - 3))/((x^2-3x)^4)`

Reducing duplicate factors yields:

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

You need to evaluate derivative at the...

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You need to find derivative of the given function using the quotient rule and the chain rule such that:

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

`f'(x) = (0*(x^2-3x)^2- 4*2(x^2-3x)(2x - 3))/((x^2-3x)^4)`

`f'(x) = (-8(x^2-3x)(2x - 3))/((x^2-3x)^4)`

Reducing duplicate factors yields:

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

You need to evaluate derivative at the point `(4,1/4)` such that:

`f'(4) = (-8*(2*4 - 3))/((4^2 - 3*4)^3)`

`f'(4) =-40/64 => f'(4) = -10/16 => f'(4) = -5/8`

**Hence, evaluating the derivative of the given function at `(4,1/4)` yields `f'(4) = -5/8.` **