
#1
Feb2612, 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 



#2
Feb2612, 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.




#3
Feb2612, 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.




#4
Feb2612, 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! 



#5
Feb2612, 03:14 AM

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P: 11,521

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? 



#6
Feb2612, 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




#7
Feb2612, 04:40 AM

HW Helper
P: 6,189

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). 



#8
Feb2612, 04:47 AM

P: 339

ok, I sort of understand.



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