What is derivative 1/(x^2-4x+3)?

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You need to differentiate the function with respect to x, using the quotient rule, such that:

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

`f'(x) = (0*(x^2-4x+3) - 1*(2x - 4))/((x^2-4x+3)^2)`

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

Factoring out 2 yields:

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

**Hence, evaluating the derivative of the function, using quotient rule, yields `f'(x) = (2(2 - x))/((x^2-4x+3)^2).` **

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

`(df)/dx= -1/2 (1/(x-3)^2 -1/(x-1)^2)` `=-1/2((x-1)^2-(x-3)^2)/[(x-1)(x-3)]^2=`

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

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