# Math Help - differentiation of simple trig fuction

1. ## differentiation of simple trig fuction

hi guys, can anyone differentiate this for me, 2sinxcosx, i'm sure it's very simple, i know how to use the chain and product rules etc, but the 2 is confusing me, do i include that in both parts? please someone show me how they would differentiate this, thanks everyone

2. ## uh oh

uh oh, this one aswell please, similar problem, 2xcos(x^2), thanks again anyone

3. Originally Posted by captainlewis
hi guys, can anyone differentiate this for me, 2sinxcosx, i'm sure it's very simple, i know how to use the chain and product rules etc, but the 2 is confusing me, do i include that in both parts? please someone show me how they would differentiate this, thanks everyone
This is product rule. The 2 is just a coefficient of the function sinxcosx. When you take the derivative of sinxcosx, the 2 stays out in front.

(2sinxcosx)'
= 2(sinxcosx)'
= 2[(sinx)'cosx + sinx(cosx)']
= 2[cos^2(x) - sin^2(x)]
= 2[cos(2x)] = 2cos(2x)

[Thanks for pointing out my mistake, captainlewis ]

4. Originally Posted by captainlewis
uh oh, this one aswell please, similar problem, 2xcos(x^2), thanks again anyone
This one involves product and chain rule.

(2xcosx^2)'
= 2(xcosx^2)'
= 2[(x)'cosx^2 + x(cosx^2)'(x^2)']
= 2[cosx^2 - xsinx^2(2x)]
= 2[cosx^2 - 2x^2*sinx^2]
= 2cosx^2 - 4x^2*sinx^2

5. ## thanks

thanks man, megamathman that was really helpful and i can do this work now, i think though cos differentiated is -sin? that's what i learnt.

6. Originally Posted by captainlewis
thanks man, megamathman that was really helpful and i can do this work now, i think though cos differentiated is -sin? that's what i learnt.
You're right, and I did that too, but I didn't "show" it. Notice how the sign in front of the term where I differentiated cosine changed from + to -.

Oops, I just notice that on the first problem I made that mistake. Sorry. I'll go back and fix it.

7. Originally Posted by captainlewis
hi guys, can anyone differentiate this for me, 2sinxcosx, i'm sure it's very simple, i know how to use the chain and product rules etc, but the 2 is confusing me, do i include that in both parts? please someone show me how they would differentiate this, thanks everyone
Here's a quicker way to do this problem:

Notice that 2sinxcosx = sin(2x) <-- double angle formula for sine.

(sin(2x))' = sin'(2x)*(2x)' = cos(2x)*2 = 2cos(2x)