Question

Greetings!
Derrivate of sin(2x+pi)
Help is much appreciated ! :)

Answer 1

the general format for a sin derivative is

d(sinx)/dx = cos(x) * d(x)/d(x)

so

d(sin(2x+pi)/dx = cos(2x+pi)*d(2x+pi) = cos(2x+pi)*2

Answer 2

here's an easier way to remember/explain it

to find the derivative of sin(a), change the sin to cos, keep the a, then multiply by the derivative of a

Answer 3

wait. think about it... what happens to the graph of sine when you add π

Answer 4

\[\sin(2x+π) = -\sin(2x)\]

Answer 5

ofc that first equation makes sense, however ur last one, I get but its hard to come to that conclusion, I understand pi jumps 180degrees making it negative, but its not obvious.

Answer 6

therefore we have:
\[\large -(\frac{d}{dx}(\sin(2x)))\]

Answer 7

chain rule it^^

Answer 8

u'v+uv'?

Answer 9

right, like so:
\[\frac{d}{dx}(\sin(2x))=\frac{dsin(u)}{du}\frac{du}{dx}\]
Im going to let \[u=2x\]
can you solve now?

Answer 10

cos(u)*2
Then u=2x so gives
Cos2x*2.
U forgot the pi tho, what happened to it?

Answer 11

as i said before, the π disappears when i rewrite the entire thing as -sin(2x)

Answer 12

and make sure you add the negative from before. The final answer is:
\[\huge -2\cos(2x)\]

Answer 13

ah okey, I will remember that, so my first step should always be to get rid of pi?
then move on the the chain rule and so forth:)
Alright, thank you :)

Answer 14

no prob. Medal please!

Answer 15

yea π is an issue

Answer 16

Haha there ya go, Always fun to learn when things go this smoothly, u explain well:)
Cheers!