Question

How do I find the derivative of 2sinxcosx

Answer 1

you have a choice
you can use the product rule of you can recall some trig identity

Answer 2

product rule sounds awesome

Answer 3

can you tell me what i'm doing wrong first?

Answer 4

second way is probably easiest,
\[2\sin(x)\cos(x)=\sin(2x)\]

Answer 5

i don't know that identity....

Answer 6

if I do 2sinxcosx
then dy/dx = 2((sinx)(-sinx) + (cosx)(cosx))

Answer 7

ok go ahead and i will check, but if the answer in the back of the book is something different, it is probably not that you did it wrong, but rather than they used some identity you don't know (aka remember)

Answer 8

it's not in the back of the book, i'm just pretty positive i don't understand this

Answer 9

no you have it correct
\[2(-\sin^2(x)+\cos^2(x))\]

Answer 10

but is is also
\[2\cos(2x)\]

Answer 11

i mean "yes, your answer is correct"

Answer 12

alright, so then an identity of cos^2x is 1-sin^2x right?

Answer 13

why is it 2cos2x haha

Answer 14

the easiest reason i can think of is that
\[\sin(2x)=2\sin(x)\cos(x)\]

Answer 15

i don't understand that though..

Answer 16

you mean why it is true?

Answer 17

yes!

Answer 18

depends on where you are willing to start, and what you know

Answer 19

if you grant that
\[\sin(\alpha +\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)\] the replace both \(\alpha\) and \(\beta\) by \(x\) and you get it in one step

Answer 20

if not, you have to use a bunch of tedious geometry to get the "addition angle" formula, and then replace \(\alpha\) and \(\beta\) by \(x\)

Answer 21

i've never heard of the addition angle. i know the product rule though

Answer 22

wait that's not the product rule

Answer 23

i know neither, i give up

Answer 24

well you already showed that you know the product rule

Answer 25

trig identities have nothing to do with calculus, they are just identities
there is a nice list here
http://math2.org/math/trig/identities.htm
i wouldn't bother memorizing them
just refer to a list when you need them

Answer 26

but you do see on that list if you look down a little that
\[\cos(2x)=\cos^2(x)-\sin^2(x)\] so your derivative, which was
\[2(\cos^2(x)-\sin^2(x))\] can be rewritten as
\[2\cos(2x)\] if you like

Answer 27

i'll keep that in mind, thank you!