# differentiation of simple trig fuction

• May 3rd 2007, 06:20 AM
captainlewis
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 :)
• May 3rd 2007, 06:29 AM
captainlewis
uh oh
uh oh, this one aswell please, similar problem, 2xcos(x^2), thanks again anyone :)
• May 3rd 2007, 07:13 AM
ecMathGeek
Quote:

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 :)]
• May 3rd 2007, 07:16 AM
ecMathGeek
Quote:

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
• May 3rd 2007, 07:30 AM
captainlewis
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.
• May 3rd 2007, 07:53 AM
ecMathGeek
Quote:

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.
• May 3rd 2007, 07:59 AM
ecMathGeek
Quote:

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)