# The derivative of ln cos x

1. Feb 26, 2012

### bobsmith76

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

Last edited: Feb 26, 2012
2. Feb 26, 2012

### 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.

3. Feb 26, 2012

### 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.

4. Feb 26, 2012

### sg001

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

5. Feb 26, 2012

### vela

Staff Emeritus
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. Feb 26, 2012

### 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

7. Feb 26, 2012

### 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).

8. Feb 26, 2012

### bobsmith76

ok, I sort of understand.

9. Feb 26, 2012

### I like Serena

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