derive [differentiate y = (-2x^2 + 5x - 1) / (2x - 1)]

[Noahb]

derive: y=(-2x^2+5x-1)/(2x-1)

i tried derving it and my answer is f'x=(-4x^2+4x-3)/(2x-1)^2
but the answer is
f'x=-1-(2/(2x-1)^2)
how can i get this ?
please help.
thank you

 

[pka]

Your answer is correct and equivalent to the given answer.
Expand your denominator and divide it out.
You will see how it works.

 

[skeeter]

\(\displaystyle y=\frac{-2x^2+5x-1}{2x-1}\)

\(\displaystyle y'=\frac{(2x-1)(-4x+5)-(-2x^2+5x-1)(2)}{(2x-1)^2}\)

\(\displaystyle y'=\frac{(-8x^2+14x-5)-(-4x^2+10x-2)}{(2x-1)^2}\)

\(\displaystyle y'=\frac{-4x^2+4x-3}{(2x-1)^2}\)

\(\displaystyle y'=\frac{-(4x^2-4x+3)}{(2x-1)^2}\)

\(\displaystyle y'=\frac{-(4x^2-4x+1)-2}{(2x-1)^2}\)

\(\displaystyle y'=\frac{-(2x-1)^2-2}{(2x-1)^2}\)

\(\displaystyle y'=\frac{-(2x-1)^2}{(2x-1)^2} - \frac{2}{(2x-1)^2}\)

\(\displaystyle y'=-1-\frac{2}{(2x-1)^2}\)

 

[ChaoticLlama]

You 'derive' equations in physics.

You 'differentiate' functions in calculus.

do not mix up the two, they are not interchangable.