Logarithmic Differentiation

[joeyjon]

Hi,

I have the answer to this problem, so I understand how to get to the final answer, but I don't understand just the first step...

f(x) = (sin x)^(tan x)

ok, you take the log of both sides -

f(x)' / f(x) = ln [(sin x)^(tan x)]

ok here is where I get confused. if the log rule for ln x^p is p ln x then why shouldn't I get...

f(x)' / f(x) = [(tan x)(sin x)]' ???


They way to correctly answer the question is for:

f(x)' / f(x) = [(tan x) + (sin x)]'


Why?

 

[Deleted member 4993]

Hi,

I have the answer to this problem, so I understand how to get to the final answer, but I don't understand just the first step...

f(x) = (sin x)^(tan x)

This is what you should get:

ln[f(x)] = tan(x) * ln[sin(x)]

f'(x)/f(x) = sec^2(x)*ln[sin(x)] + tan(x)*tan(x) = sec^2(x) * {1 + ln[sin(x)]} - 1


ok, you take the log of both sides -

f(x)' / f(x) = ln [(sin x)^(tan x)]

ok here is where I get confused. if the log rule for ln x^p is p ln x then why shouldn't I get...

f(x)' / f(x) = [(tan x)(sin x)]' ???


They way to correctly answer the question is for:

f(x)' / f(x) = [(tan x) + (sin x)]'


Why?

.

 

[HallsofIvy]

Don't you think it would have been a good idea to tell us what the problem really was? I guess that you mean to find the derivative of y with respect to x but you never actually say that!