How do you differentiate #y= log _a x#?

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Answer 1 · 2016-07-22T18:14:33

#y' = 1/(xln a)#

the easiest way is to shift the base to e

so #y = log_a x = (log_e x)/(log_e a)# {small demo of what that is so is set out below}

thusly

#y' = 1/x *1/(log_e a) = 1/(xln a)#

the demo

#y = log_a x implies a^y = x# by definition

so we choose to use natural logs because they work so well with calculus

#ln a^y = ln x#

#y ln a = ln x#

#y = ( ln x)/(ln a)#