Understanding the Derivative of ln(cos x) for Plane Curves?

[bobsmith76]

Homework Statement

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

 

[intwo]

Because $\ln{\cos{x}}$ is actually a composition of the functions $\ln{x}$ and $\cos{x}$. Therefore you need to use the chain rule.

 

[bobsmith76]

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]

you need to use the derivative of ln r(t), which is r'(t)/ r (t)
try that!

 

[vela]

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]

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

 

[I like Serena]

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]

ok, I sort of understand.

 

[I like Serena]

ok, I sort of understand.

So do you understand how the book got -tan(t)?