- #1
Dustobusto
- 32
- 0
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.
Find the derivative:
y = ln(tan x)
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)
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?
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?