How do you differentiate #f(x)=Log_5(x)#?

Question

1420 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-24T03:45:28

#(df)/(dx)=0.6213/x#

WE can use here the identity #log_ac=log_abxxlog_bc#

hence #log_5x=log_5exxlog_ex#

= #loge/log5xxlnx#

Hence #f(x)=loge/log5xxlnx#

Now as #loge/log5# is a constant term,

#(df)/(dx)=loge/log5xx1/x#

= #0.4343/0.69897xx1/x=0.6213/x#