change the base to ln
remmber
[tex]log_a(b)= \frac{log_c(b)}{log_c(a)} [/tex]
ln is just base e
so
[tex]log_5(2x^3-1)= \frac{log_e(2x^3-1)}{log_e(5)} = \frac{ln(2x^3-1))}{ln(5)} [/tex]
remember that tthe deritivive of [tex] \frac{f(x)}{g(x)} [/tex] is [tex] \frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2} [/tex]
so
let's take the deritivies seperately first
deritivive of ln(2x^3-1) is [tex] \frac{6x^2}{2x^3-1} [/tex]
deritivive of ln(5) is 0
[tex]\frac{ln(2x^3-1))}{ln(5)} = \frac{( \frac{6x^2}{2x^3-1})(ln(5))-(0)(ln(2x^3-1))}{(ln(5))^2} [/tex]=[tex] \frac{( \frac{6x^2}{2x^3-1})(ln(5))}{(ln(5))^2}[/tex]=[tex]\frac{ \frac{6x^2}{2x^3-1}}{ln(5)}= \frac{6x^2}{(2x^3-1)(ln(5))}[/tex]
