Question

Find the derivatives of the following functions.
${\log }_2(2x^2-3x+1)$

Answer 1

$f\left(x\right)={\log }_2(2x^2-3x+1)$

$=\frac{\log 2x^2-3x+1}{\log 2}$

$\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left[\frac{\log 2x^2-3x+1}{\log 2}\right]$

$=\frac{1}{\log 2}\frac{d}{dx}[\log 2x^2-3x+1]$

$=\frac{1}{\log 2}\left[\frac{1}{2x^2-3x+1}\frac{d}{dx}\right(2x^2-3x+1\left)\right]$

$=\frac{1}{\log 2}\left[\frac{4x-3}{2x^2-3x+1}\right]=\frac{4x-3}{\left(\log 2\right)(x-2)(x-1)}$.