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## anonymous one year ago Greetings! Derrivate of sin(2x+pi) Help is much appreciated ! :)

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1. Vocaloid

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

2. Vocaloid

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

3. anonymous

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

4. anonymous

$\sin(2x+π) = -\sin(2x)$

5. anonymous

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.

6. anonymous

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

7. anonymous

chain rule it^^

8. anonymous

u'v+uv'?

9. anonymous

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?

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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 :)

14. anonymous

no prob. Medal please!

15. anonymous

yea π is an issue

16. anonymous

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

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