# Simplifying a derivative

1. Oct 11, 2008

### DollarBill

1. The problem statement, all variables and given/known data

Find the derivative of
$$f(x)=2x^3+cos^2(x^3)$$

3. The attempt at a solution

$$f'(x)=6x^2+2cos(x^3)*-sin(x^3)*3x^2$$
$$f'(x)=6x^2-6x^2cos(x^3)sin(x^3)$$

The book has a different solution since they always simplify, but how would I simplify it? I know it has something to do with the 6x2, but the 2nd 6x2 is attached to the cos.

2. Oct 11, 2008

### Defennder

Just use the double angle formula for sin. What is sin2x for example?

3. Oct 11, 2008

### DollarBill

Never heard of it

4. Oct 11, 2008

### gabbagabbahey

It's a Trig Identity that you should have come across in high school:

$$sin(2\theta)=2sin(\theta)cos(\theta)$$

5. Oct 11, 2008

### DollarBill

I'm still in high school...

6. Oct 11, 2008

### HallsofIvy

Staff Emeritus
Okay, so factor it out:
$$f'(x)= 6x^2(1- cos(x^3)sin(x^3))$$
You don't say WHAT the "different solution" in the book is so I don't know if you want that simplified more. You could use the identity Defennder gave you: Since sin(2x3)= 2sin(x3)cos(x3, cos(x3)sin(x3= (1/2)sin(2x3).

Whether you are in High School or not, taking a course before learning the pre-requisites for that course is just wasting your time. And trigonometry is definitely a pre-requisite for problems like this.

7. Oct 11, 2008

### DollarBill

I've taken PreCalc. I just don't recall learning that identity. Maybe I just forgot or it was called by a different name. The only identities that I really remember well is the basic Pythagorean identity.

Thanks for the help though.