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Study23 Group Title

{Derivatives, Focus: Chain Rule. Help appreciated} \(\ \Huge y=cos(a^3+x^3) . \)

  • 2 years ago
  • 2 years ago

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  1. Dido525 Group Title
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    Is a a constant?

    • 2 years ago
  2. campbell_st Group Title
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    are you differentiating with respect to x..?

    • 2 years ago
  3. Dido525 Group Title
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    or a? :P .

    • 2 years ago
  4. Study23 Group Title
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    Yes @campbell_st. I'm not sure @Dido525

    • 2 years ago
  5. Dido525 Group Title
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    That means a is a constant I assume O_o .

    • 2 years ago
  6. Study23 Group Title
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    \(\ \text{ I got to this point, but my text tells me this answer is incorrect: } y'=-sinx(3a^2+3x^2). \)

    • 2 years ago
  7. Dido525 Group Title
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    |dw:1352349392745:dw|

    • 2 years ago
  8. campbell_st Group Title
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    let u = a^3 + x^3 du/dx = 3x^2 let y = cos(u) du/dy = -sin (u) so dy/dx = dy/du * du/dx

    • 2 years ago
  9. Dido525 Group Title
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    I can't see the rest.

    • 2 years ago
  10. Study23 Group Title
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    One moment

    • 2 years ago
  11. Study23 Group Title
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    \(\ \Huge -sinx(3a^2+3x^2) \)

    • 2 years ago
  12. campbell_st Group Title
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    then replace u with a^3 + x^3 for the final solution.

    • 2 years ago
  13. campbell_st Group Title
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    nope its dy/dx = -sin(a^3 + x^3) * 3x^2

    • 2 years ago
  14. Study23 Group Title
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    So the way my math teacher did this in class was using @Dido525 illustrated

    • 2 years ago
  15. Study23 Group Title
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    @campbell_st How did you get that?

    • 2 years ago
  16. Dido525 Group Title
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    Because a is a constant. The derivative of a constant is 0.

    • 2 years ago
  17. Dido525 Group Title
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    We multiply by the derivative of the inside part when using the chain rule.

    • 2 years ago
  18. campbell_st Group Title
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    and I used the chain rule..

    • 2 years ago
  19. aerokat25 Group Title
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    @camp is correct per calculus instructor

    • 2 years ago
  20. campbell_st Group Title
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    the a^3 x^3 doesn't change

    • 2 years ago
  21. Study23 Group Title
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    @campbell_st I thought a was a constant then? How does that stay the same?

    • 2 years ago
  22. Dido525 Group Title
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    Here: |dw:1352349668500:dw|

    • 2 years ago
  23. Dido525 Group Title
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    We DO NOT change the inside.

    • 2 years ago
  24. aerokat25 Group Title
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    Now if you are implicitly differentiating then you could possibly get a different answer if a is not constant. However I'm pretty sure a is a constant in this case

    • 2 years ago
  25. Study23 Group Title
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    |dw:1352349758941:dw|

    • 2 years ago
  26. Dido525 Group Title
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    Differentiate that.

    • 2 years ago
  27. Dido525 Group Title
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    With respect to x. Meaning a is a constant. We know the derivative of a constant is 0.

    • 2 years ago
  28. thewilled1 Group Title
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    Usually letters a, b, and c are considered constants.

    • 2 years ago
  29. Study23 Group Title
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    So you can't differentiate more than one variable in any given function? Only variable is differentiated in a problem with more than one variable?

    • 2 years ago
  30. aerokat25 Group Title
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    Also check to see if you typed the question correctly because that answer you typed from the back of the book looks like you changed the angle measurement to x

    • 2 years ago
  31. Dido525 Group Title
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    Well there are partial derivatives but you probably won't learn that now.

    • 2 years ago
  32. Study23 Group Title
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    @aerokat25 The correct answer the book provides is \(\ y'=-3x^2sin(a^3+x^3) .\) Why can't I differentiate x^3 ?

    • 2 years ago
  33. campbell_st Group Title
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    its just like (x^2 + 3)^4 dy/dx = 4(x^2 + 3) ^3 *2x the brackets don't change

    • 2 years ago
  34. campbell_st Group Title
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    you do... thats how you get 3x^2

    • 2 years ago
  35. Dido525 Group Title
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    When you have a function Just inside another function the function inside the brackets DO NOT change.

    • 2 years ago
  36. Study23 Group Title
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    But the term is still inside the parenthesis? That's really confusing me right now

    • 2 years ago
  37. Study23 Group Title
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    parentheses*

    • 2 years ago
  38. campbell_st Group Title
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    that is why textbooks talk about u = a^3 + x^3 du/dx = 3x^2 y = cos(u) dy/du = -sin(u) dy/dx = dy/du * du/dx

    • 2 years ago
  39. Dido525 Group Title
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    |dw:1352350103544:dw|

    • 2 years ago
  40. aerokat25 Group Title
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    Chain rule says d/dx(cos u)= -sin u (u')

    • 2 years ago
  41. campbell_st Group Title
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    yep and du/dx = u' you are differentiating u with respect to x

    • 2 years ago
  42. Study23 Group Title
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    Hmmm. That's really confusing because our teacher shows us the u method on one example, and said he doesn't use it. So I don't really follow, to be honest.

    • 2 years ago
  43. aerokat25 Group Title
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    So u= a^3 + x^3

    • 2 years ago
  44. aerokat25 Group Title
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    U'= 0 + 3x^2

    • 2 years ago
  45. campbell_st Group Title
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    well its the derivative of the outer function cos times the derivative of the inner function a^3 + x^3

    • 2 years ago
  46. Study23 Group Title
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    Okay, I get that part, but the one thing I am not understanding is how the expression remains. I'm sorry if I'm being silly or something.

    • 2 years ago
  47. aerokat25 Group Title
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    Ok well u know d/dx of cos x= -sin x right?

    • 2 years ago
  48. Dido525 Group Title
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    The expression inside of the expression on the outside ALWAYS remains. It's just how it is.

    • 2 years ago
  49. Study23 Group Title
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    \(\ \text{Yes!} \)

    • 2 years ago
  50. Study23 Group Title
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    "on the outside"?

    • 2 years ago
  51. aerokat25 Group Title
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    Well d/dx cos x actually equals -sin x (x') but the derivative of x is just 1 so you have actually been using chain all along and didn't really know it.

    • 2 years ago
  52. Study23 Group Title
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    Okay\(\ \text{...} \)

    • 2 years ago
  53. aerokat25 Group Title
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    And that angle measure x has never changed right? So you aren't going to change what you're given as an angle measurement now either. You just have to multiply by its derivative which isn't going to be a 1 anymore

    • 2 years ago
  54. Study23 Group Title
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    Angle measure?

    • 2 years ago
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