Question

How do you find the derivative of `f(x) = log_x (3)`?

Answer 1

`(df)/(dx)(x) = -log_e3/(x log_e^2(x))`

Explanation:

`y = log_x 3 = (log_b 3)/(log_b x)` for any convenient basis `b`

Calling now `g(x,y) = y log_e x - log_e 3 = 0` after taking `b = e`, we have

`dg = g_x dx + g_y dy = 0`

then

`(dy)/(dx) = - (g_x)/(g_y) = -((y/x))/(log_e x) = -y/(x log_e(x)) = -log_e3/(x log_e^2(x))`