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

1 Answer
Cesareo R.
Jun 1, 2016

#(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))#

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