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

Question

19599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-19T12:55:01

#(dy)/(dx)=1/(xlnb)#

Before differentiating we need to change the log to base e

#y=log_bx#

#=>b^y=x#

take logs to base e

#lnb^y=lnx#

#=>ylnb=lnx#

now differentiate wrt #x#

#d/(dx)(ylnb)=d/(dx)(lnx)#

#lnb(dy)/(dx)=1/x#

#(dy)/(dx)=1/(xlnb)#