How to Simplify a Derivative with Multiple Terms

[DollarBill]

Homework Statement

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

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.

 

[Defennder]

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

 

[DollarBill]

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

Never heard of it

 

[gabbagabbahey]

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

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

 

[DollarBill]

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

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

I'm still in high school...

 

[HallsofIvy]

Homework Statement

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

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.

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.

 

[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.