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Study23

  • 2 years ago

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

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  1. Dido525
    • 2 years ago
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    Is a a constant?

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

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

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

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

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

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

  8. campbell_st
    • 2 years ago
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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

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

  10. Study23
    • 2 years ago
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    One moment

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

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

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

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

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

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

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

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

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

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

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

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

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

  24. aerokat25
    • 2 years ago
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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

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

  26. Dido525
    • 2 years ago
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    Differentiate that.

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

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

  29. Study23
    • 2 years ago
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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?

  30. aerokat25
    • 2 years ago
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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

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

  32. Study23
    • 2 years ago
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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 ?

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

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

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

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

  37. Study23
    • 2 years ago
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    parentheses*

  38. campbell_st
    • 2 years ago
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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

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

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

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

  42. Study23
    • 2 years ago
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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.

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

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

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

  46. Study23
    • 2 years ago
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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.

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

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

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

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

  51. aerokat25
    • 2 years ago
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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.

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

  53. aerokat25
    • 2 years ago
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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

  54. Study23
    • 2 years ago
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    Angle measure?

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