Question

How do you find the derivatives of `y=log_x(a)`?

Answer 1

`d/(dx) (log_x(a)) = -lna/(xln^2(x))`

Explanation:

You can use this property of logarithms:

`log_b(a) = lna/lnb`

so that:

`y= lna/lnx`

and:

`(dy)/(dx) = d/(dx) (log_x(a))= d/(dx) (lna/lnx) = lna*1/x*(-1/ln^2(x))= -lna/(xln^2(x))`