the derivative of ln cos x

by bobsmith76
Tags: derivative
 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
 P: 107 Because $\ln{\cos{x}}$ is actually a composition of the functions $\ln{x}$ and $\cos{x}$. Therefore you need to use the chain rule.
 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.
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!
 PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,862 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?
 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
 HW Helper P: 6,164 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).
 P: 339 ok, I sort of understand.
HW Helper
P: 6,164
 Quote by bobsmith76 ok, I sort of understand.
So do you understand how the book got -tan(t)?

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