Question

How do you differentiate `f(x)=(2x^2-x+1)/(2x-1)` using the quotient rule?

Answer 1

`f^'(x)=[4x^2-4x-1]/[(2x-1)^2]`

Explanation:

Quotient rule states that if `h` and `k` are two differentiable functions on `]a,b[`, then `forall x in ]a,b[` such that `k(x) ne 0` the following equality holds:
`[(h(x))/(k(x))]^'=[h^'(x)k(x)-h(x)k^'(x)]/(k^2(x))`

In our case `h(x)=2x^2-x+1` and `k(x)=2x-1`. The two derivatives are `h^'(x)=4x-1` and `k^'(x)=2`.
`f^'(x)=[(4x-1)(2x-1)-(2x^2-x+1)2]/[(2x-1)^2]=[8x^2-6x+1-4x^2+2x-2]/[(2x-1)^2]=[4x^2-4x-1]/[(2x-1)^2]`