Given
[tex]y = \log_5\bigg((2x + 1) (x - 3)\bigg)[/tex]
Expand the logarithm on the right side. (change-of-base and product-to-log identities)
[tex]y = \dfrac{\ln(2x + 1) + \ln(x - 3)}{\ln(5)}[/tex]
Now differentiate both sides with respect to [tex]x[/tex].
[tex]y' = \dfrac1{\ln(5)} \bigg(\ln(2x+1)\bigg)' + \dfrac1{\ln(5)} \bigg(\ln(x-3)\bigg)'[/tex]
[tex]y' = \dfrac{(2x+1)'}{\ln(5)\,(2x+1)} + \dfrac{(x-3)'}{\ln(5)\,(x-3)}[/tex]
[tex]y' = \dfrac{2}{\ln(5)\,(2x+1)} + \dfrac{1}{\ln(5)\,(x-3)}[/tex]
[tex]y' = \boxed{\dfrac{4x-5}{\ln(5)\,(2x+1)(x-3)}}[/tex]
