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anonymous

  • one year ago

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

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  1. Vocaloid
    • one year ago
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    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
    • one year ago
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    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. Rizags
    • one year ago
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    wait. think about it... what happens to the graph of sine when you add π

  4. Rizags
    • one year ago
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    \[\sin(2x+π) = -\sin(2x)\]

  5. anonymous
    • one year ago
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    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. Rizags
    • one year ago
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    therefore we have: \[\large -(\frac{d}{dx}(\sin(2x)))\]

  7. Rizags
    • one year ago
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    chain rule it^^

  8. anonymous
    • one year ago
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    u'v+uv'?

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

  11. Rizags
    • one year ago
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    as i said before, the π disappears when i rewrite the entire thing as -sin(2x)

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

  13. anonymous
    • one year ago
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    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. Rizags
    • one year ago
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    no prob. Medal please!

  15. Rizags
    • one year ago
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    yea π is an issue

  16. anonymous
    • one year ago
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    Haha there ya go, Always fun to learn when things go this smoothly, u explain well:) Cheers!

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