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The derivative of ln cos x

by bobsmith76
Tags: derivative
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bobsmith76
#1
Feb26-12, 02:12 AM
P: 339
1. The problem statement, all variables and given/known data

the question is for plane curves. find T N and k for the plane curves

1. r(t) = ti + ln (cos t)

the derivative of ln x is 1/|x|, so why isn't the derivative of ln cos x, 1/cos x or sec x?

My book says the answer is - tan x
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intwo
#2
Feb26-12, 02:45 AM
P: 107
Because [itex]\ln{\cos{x}}[/itex] is actually a composition of the functions [itex]\ln{x}[/itex] and [itex]\cos{x}[/itex]. Therefore you need to use the chain rule.
bobsmith76
#3
Feb26-12, 02:52 AM
P: 339
I put in the question into the original post. I don't see how they're using the chain rule. They aren't deriving two terms, it's just ln (cos t). if they're driving t, then i would think the derivative of t would just be 1.

sg001
#4
Feb26-12, 02:56 AM
P: 134
The derivative of ln cos x

you need to use the derivative of ln r(t), which is r'(t)/ r (t)
try that!
vela
#5
Feb26-12, 03:14 AM
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Consider this problem: Find the derivative of ##y=\ln t^2##.

First method: Using the property of logs, we pull the exponent out front.
$$y=\ln t^2 = 2\ln t$$ When you differentiate this, you get
$$y' = 2\left(\frac{1}{t}\right).$$ This is the correct answer.

Second method: Just "differentiate" the way you did
$$y' = \frac{1}{t^2}.$$ This is wrong. Why is it wrong?
bobsmith76
#6
Feb26-12, 04:27 AM
P: 339
but there's no exponent involved here. plus look at this problem: they just derive everything normally here. they don't do the r'/r

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#7
Feb26-12, 04:40 AM
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The chain rule says that the derivative of f(g(t)) is f'(g(t))g'(t).
This is what vela illustrated.

In your case, for ln(cos(t)), you have calculated f'(g(t)) which is 1/cos(t), but you still need to multiply by g'(t) which is the derivative of cos(t).
bobsmith76
#8
Feb26-12, 04:47 AM
P: 339
ok, I sort of understand.
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#9
Feb26-12, 04:54 AM
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Quote Quote by bobsmith76 View Post
ok, I sort of understand.
So do you understand how the book got -tan(t)?


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