Differentiate y = log e (log e (x))
Here e is base of log.
kvnmurty:
is that Log_e (log_e x ) ? where e is the base for both Log ?
or Log_e [ Log e^x] ? e power x ??
what is correct ?
Answers
Answered by
3
dy/dx=d(loge(log e (x))/dx
=1/log e (x) d log e (x)/dx
=1/x log e (x)
=1/log e (x) d log e (x)/dx
=1/x log e (x)
Your answer is wrong
oh sorry
The answer is correct if the question is : y = Log [ Log x ] ... here e is treated as the base...
the qn is not clear
Answered by
1
we also write loge = ln
use this,
y = ln( ln(e^x))
differentiate wrt x
dy/dx = 1/(ln(e^x) d(lne^x)/dx
= 1/ln(e^x) × 1/e^x × d(e^x)/dx
= 1/ln(e^x) × e^x/e^x
=1/ln(e^x )
use this,
y = ln( ln(e^x))
differentiate wrt x
dy/dx = 1/(ln(e^x) d(lne^x)/dx
= 1/ln(e^x) × 1/e^x × d(e^x)/dx
= 1/ln(e^x) × e^x/e^x
=1/ln(e^x )
1/ Ln (e^x) is simply 1/x
is the qn : y = Ln {ln e^x) ? it is simply y = Ln x as Ln e^x = x itself.
yeah !!! right
Similar questions
Science,
8 months ago
Hindi,
8 months ago
Physics,
1 year ago
Political Science,
1 year ago
English,
1 year ago
Social Sciences,
1 year ago
Science,
1 year ago