Question

Differentiate.

y = log₅(2x³-1)

Answer 1

change the base to ln

remmber

`log_a(b)= (log_c(b))/(log_c(a))`
ln is just base e

so

`log_5(2x^3-1)= (log_e(2x^3-1))/(log_e(5)) = (ln(2x^3-1)))/(ln(5))`

remember that tthe deritivive of
`(f(x))/(g(x))` is
`(f'(x)g(x)-g'(x)f(x))/(g(x)^2)`
so

let's take the deritivies seperately first
deritivive of ln(2x^3-1) is
`(6x^2)/(2x^3-1)`
deritivive of ln(5) is 0



`(ln(2x^3-1)))/(ln(5)) = (( (6x^2)/(2x^3-1))(ln(5))-(0)(ln(2x^3-1)))/((ln(5))^2)`=
`(( (6x^2)/(2x^3-1))(ln(5)))/((ln(5))^2)`=
`( (6x^2)/(2x^3-1))/(ln(5))= (6x^2)/((2x^3-1)(ln(5)))`

Answer 2

(1/(ln 5))((6x^2)/(2x^3-1))