Derivatives of General Exponential and Logarithmic Functions

[Dustobusto]

So in my math class we're studying derivatives involving ln(), tanh, coth, etc..

I need to say this first. I skipped precalc and trig and went straight to calculus, so whenever I see a trig problem, I can only go off of what I've learned "along the way." This problem has baffled me, please help me out. It seems rather simple in nature so it shouldn't take too long to solve this.

Homework Statement

Find the derivative:

y = ln(tan x)

Homework Equations

So there are a ton of rules involving ln() functions. Here's a couple

The derivative (d/dx) of ln(x) = 1/x
d/dx of ln[f(x)] = derivative of f(x) over f(x) or f'(x)/f(x)

The Attempt at a Solution

So, I learned that in these scenarios, tan x, sec x, sin x, etc. are considered composite functions. So I used f'(x)/f(x) to solve.

f(x) is clearly tan x. The book says the derivative of tan x = sec² x. So I end up with, as my answer

sec² x/ tan x.

The back of the book gives 1/(sin x cos x)

Am I missing a trigonometric rule here? Did I perform this incorrectly?

 

[micromass]

You did it perfectly. All you need to do now is to show that

[tex]\frac{sec^2(x)}{\tan(x)} = \frac{1}{\sin(x)\cos(x)}[/tex]

Try to prove this by expressing both sec and tan in terms of sin and cos.

 

[Dustobusto]

ok nice to know