# Differentiating trigonometric F'ns with double angles

1. Mar 9, 2013

### Feodalherren

1. The problem statement, all variables and given/known data
Sin(tan(2x))
With respect to x

2. Relevant equations
Differentiation

3. The attempt at a solution

My question is whether I can simply use d/dx (Tan x) = Sec^2(X) to extrapolate that to d/dx(tan 2x) = Sec^2(2x) ?

Or do I have to convert to sine/cosine and go from there?

2. Mar 9, 2013

### MarneMath

Well,

You I want to point out that (tan2x)' does not equal to sec^2(2x). (Error in technique).

However, the generally idea isn ok (assuming you using the chain rule in regards to sin(tan(2x)).

3. Mar 9, 2013

### Feodalherren

So this problem can't be solved by using the chain rule thusly:

d/dx Sin(tan 2x) = cos(tan2x)Sec^2(2x)?

It turns into:

Sin(tan 2x) (2)d/dx [sinxcosx / (cos^2(X) - Sin^2(x))]

quotient rule etc... damn this is gonna be messy. Am I on the right track at least?

Last edited: Mar 10, 2013
4. Mar 10, 2013

### MarneMath

"d/dx Sin(tan 2x) = cos(tan2x)Sec^2(2x)?"

This is essentially correct, but you're making an error when you take the derivative of tan2x. I'm hoping you look at that part just a little harder and figure out what that error exactly is.

5. Mar 10, 2013

### Feodalherren

d/dx tan 2x = [sec^2 (2x)] (2)

?