How do you find the derivative of #log_10 x#?

Question

9815 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-23T22:17:04

See below

Maybe learn how to base shift

Let #y = log_a b color(blue)( implies a^y = b)#

And for new base #Q#:

#log_Q a^y= log_Q b implies y log_Q a= log_Q b#

So:

#y = color(red)( (log_Q b)/(log_Q a) = log_a b)# <-- base-shifting

Pattern matching:

#d/(dx) (log_(10) x) #

#= d/(dx)( (log_e x)/(log_e 10) ) = d/(dx)( (ln x)/(ln 10) ) #

# = (1)/(x ln 10) #

PS: Assumes you know that #(ln x)' = 1/x#