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Christos

Find derivative of http://screencast.com/t/sXZqTgJKM Just tell me 1-2 steps I will clip to them afterwards

  • 10 months ago
  • 10 months ago

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  1. abb0t
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    product rule within chain rule.

    • 10 months ago
  2. zepdrix
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    \[\large y=x^3 \sin^2(5x)\]We start by setting up the product rule.\[\large y'=\color{royalblue}{(x^3)'}\sin^2(5x)+x^3\color{royalblue}{(\sin^2(5x))'}\]

    • 10 months ago
  3. Christos
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    I already got to that point @zepdrix that's where I am stuck :D 2x^2sin(5x)+2x^3sin(5x)

    • 10 months ago
  4. Christos
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    I forgot a power of 2 on the first sin *

    • 10 months ago
  5. zepdrix
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    \[\large y'=\color{orangered}{3x^2}\sin^2(5x)+x^3\color{royalblue}{(\sin^2(5x))'}\]I've taken the derivative of the first one. I think your coefficient is a lil mixed up. You brought the wrong exponent down :O

    • 10 months ago
  6. Christos
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    Bro which one are you referring to

    • 10 months ago
  7. zepdrix
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    The first blue term in the original product rule that I wrote out.

    • 10 months ago
  8. Christos
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    That? Am I not supposed to take down the power and multiply it with whatever it is next to it and and substract the power by 1?

    • 10 months ago
  9. zepdrix
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    x^3. Bringing the power down and subtracting by one gives us, 3x^2 right? :o

    • 10 months ago
  10. Christos
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    I am talking about the power of sin not x

    • 10 months ago
  11. Christos
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    Remember now we differantiating the second part of the expression or whatever

    • 10 months ago
  12. zepdrix
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    I wasn't talking about that one. I was doing the other term.

    • 10 months ago
  13. zepdrix
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    The second one will be a lil more complicated, I wanted to make sure you got the first part.

    • 10 months ago
  14. Christos
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    ah yea you got a point :D That one I didnt notice :D ok

    • 10 months ago
  15. zepdrix
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    So for the second one, again we'll apply the power rule to the sine function, then we'll apply the chain rule, multiplying by the derivative of the inside. \[\large y'=\color{orangered}{3x^2}\sin^2(5x)+x^3\color{orangered}{(2\sin(5x))}\color{royalblue}{(\sin(5x))'}\]The new blue term that showed up is the one we need to take a derivative of. If that is confusing, there is another way we can write it.

    • 10 months ago
  16. Christos
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    What we do this only for the second part if I may ask

    • 10 months ago
  17. Christos
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    Just need to understand whats the difference that forces us to do so

    • 10 months ago
  18. Christos
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    why*

    • 10 months ago
  19. zepdrix
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    So we start with the product of two things involving x. \[\large y=(x^3)(\sin^2(5x))\] Without taking any derivatives yet, we can setup the product rule. \[\large y'=\color{royalblue}{(x^3)'}\sin^2(5x)+x^3\color{royalblue}{(\sin^2(5x))'}\] The blue terms are the ones we have to differentiate (take derivatives of). So you can see in our setup, we only differentiate the x^3 in the first term, and the sine in the second term. Was that the question? :o It's because of the product rule.

    • 10 months ago
  20. Christos
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    No I mean after that we do we take chain rule for Only the second term? Chain rule appears to be a lil messed up in my mind

    • 10 months ago
  21. zepdrix
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    Yah the second blue term has an inner function. Umm I'm trying to think of a good way to explain it...

    • 10 months ago
  22. Christos
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    if I have this (y5x)^2 Will I take chain rule for derivative?

    • 10 months ago
  23. Christos
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    (5x)^2 *** ignore the fact that I can just do it 25x^2 lets say I COULDNT. Will I use chain rule?

    • 10 months ago
  24. zepdrix
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    Yes that's a very good example :) Here's how we would work it out.\[\large \left[(5x)^2\right]' \qquad = \qquad 2(5x)\]That's the derivative of the OUTER FUNCTION. (Something squared). Now we have to multiply by the derivative of the inside.\[\large \left[(5x)^2\right]' \qquad = \qquad 2(5x)(5x)' \qquad = \qquad 2(5x)(5)\]

    • 10 months ago
  25. zepdrix
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    The chain rule can be really tricky to get a handle on. Lemme know if you need another example.

    • 10 months ago
  26. Christos
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    I see :) Alright I got it. Just something last because according to what I learned now I still cant make the solution like that in the solution manual (unofficial manual) http://screencast.com/t/snYsnYAbha maybe the above link is wrong?

    • 10 months ago
  27. zepdrix
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    No those steps are correct but they made a substitution to try and get the point across. I find that more confusing personally.. hmm

    • 10 months ago
  28. Christos
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    My own solution is 3x^2sin^2(5x)+2x^3sin(5x)cos(5x)

    • 10 months ago
  29. Christos
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    oh sh*t there is a 5 too hold on

    • 10 months ago
  30. Christos
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    3x^2sin^2(5x)+10x^3sin(5x)cos(5x)

    • 10 months ago
  31. zepdrix
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    Yes very good :) So you applied the chain rule TWICE, yes? After you got a cosine, you took the derivative of the (5x)

    • 10 months ago
  32. Christos
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    yea

    • 10 months ago
  33. Christos
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    but still the manual doesn't end up with any cos

    • 10 months ago
  34. zepdrix
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    What they did in the solutions manual is, they applied the `Double Angle Formula for Sine`. That's why it looks a bit different.\[\large 2\sin x \cos x = \sin 2x\]So if you look at your second term, ignoring the x^2 in front, we have something like this,\[\large 5\cdot2 \sin(5x)\cos(5x)\]Applying the double angle formula gives us,\[\large 5 \cdot \sin(10x)\]

    • 10 months ago
  35. zepdrix
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    They did have a cosine if you look at the middle steps. They applied an annoying trig rule to get rid of it though :)

    • 10 months ago
  36. Christos
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    ooh so thats why

    • 10 months ago
  37. Christos
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    Not important for my course, im glad my way is correct too, thanks again

    • 10 months ago
  38. zepdrix
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    np c:

    • 10 months ago
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