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anonymous

  • 5 years ago

Find dy/du, du/dx, and dy/dx y= u^2 , u=4x+7

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  1. anonymous
    • 5 years ago
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    2u,4,8u in order as give. :)

  2. anonymous
    • 5 years ago
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    \[\frac{dy}{du}=2u\]\[\frac{du}{dx}=4\]\[\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=4\times2u=8u\]

  3. anonymous
    • 5 years ago
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    can u guys show the working because i dont understand how u got that did u use the quotient rule to find dy/du

  4. anonymous
    • 5 years ago
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    purple did you get and understand the solution ?

  5. anonymous
    • 5 years ago
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    no

  6. anonymous
    • 5 years ago
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    Just use the normal rule for differentiating polynomials to find dy/du and du/dx and the use the chain rule to find dy/dx.

  7. anonymous
    • 5 years ago
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    ok

  8. anonymous
    • 5 years ago
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    ok i cant do it for you dy/dx =2u ,du/dx==4 and dy/dx= (dy/du)(du/dx)=2u(4)= 8u

  9. anonymous
    • 5 years ago
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    wouldnt dy/dx be 8(4x+7)

  10. anonymous
    • 5 years ago
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    i mean i can do it for u..lol..

  11. anonymous
    • 5 years ago
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    using the chain rule

  12. anonymous
    • 5 years ago
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    dy/dx =2u ,du/dx==4 and dy/dx= (dy/du)(du/dx)=2u(4)= 8u

  13. anonymous
    • 5 years ago
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    did you understand my solution purple?

  14. anonymous
    • 5 years ago
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    we didnt learn to do it like that we do f'(x) = f'(g(x))g'(x)

  15. anonymous
    • 5 years ago
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    ok that is the same...lol.isnt that dy/dx is very simple?

  16. anonymous
    • 5 years ago
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    nope its 8(4x+7)

  17. anonymous
    • 5 years ago
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    @ least its how i learned how to do it

  18. anonymous
    • 5 years ago
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    dy d(u^2) du d(4x+7) ---- = -----=2u , --- = ------ 4 du du dx dx

  19. anonymous
    • 5 years ago
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    du/dx =4

  20. anonymous
    • 5 years ago
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    now dy dy du ---= --- ----= 2u(4)= 8u ans.. dx du dx

  21. anonymous
    • 5 years ago
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    is it much better now purple?

  22. anonymous
    • 5 years ago
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    no its the same...i'll just do that one the way i learned and see what my lectuer says

  23. anonymous
    • 5 years ago
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    i didnt see that bendt did it up there..lol bec my pc keep hanging up..lol

  24. anonymous
    • 5 years ago
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    but you problem reqiured dy/dx, du/dx and dy/du instead of using f(x)

  25. anonymous
    • 5 years ago
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    ok

  26. anonymous
    • 5 years ago
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    ok if you want to use functions dy/du=f '(u)=2u du/dx=f '(x)=4 dy/dx=f '(u) f '(x)= 2u(4)=8u

  27. anonymous
    • 5 years ago
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    hope you get it now?..lol

  28. anonymous
    • 5 years ago
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    yes

  29. anonymous
    • 5 years ago
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    thank you so much

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