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DLS

  • 3 years ago

Find dy/dx !

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  1. DLS
    • 3 years ago
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    \[y=\cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c} \]

  2. DLS
    • 3 years ago
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    I want to build my logic and way of attacking problems !

  3. DLS
    • 3 years ago
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    \[\large y= \cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c}\]

  4. MarcLeclair
    • 3 years ago
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    errr implicit derivative... only with variables? D:

  5. MarcLeclair
    • 3 years ago
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    no actual points? xD its going to get messy

  6. DLS
    • 3 years ago
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    haha,i know..and no sorry :D

  7. MarcLeclair
    • 3 years ago
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    alright i'll try it out, I'm taking cal 1, good review for my test. Hum first, the whole thing is an addition of 2 big terms, so find the derivative of bot terms. The first one is a chain rule : -sin s(ax ^2+bx+c) * (2ax + b)

  8. DLS
    • 3 years ago
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    yes!! i did that

  9. DLS
    • 3 years ago
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    what about the sin^3 part :S

  10. MarcLeclair
    • 3 years ago
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    secondly comes the other term. Because its a root we'll simplify it to: sin ^3 (ax ^2+bx+c)^(1/2) Again the chain rule So we get cos^3(ax ^2+bx+c)^(1/2) * (ax ^2+bx+c)^(1/2)' Therefore we have ANOTHER chain rule: 1/2 (ax ^2+bx+c)^-(1/2) * (2ax+b)

  11. MarcLeclair
    • 3 years ago
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    so you have what I think of a chainception

  12. MarcLeclair
    • 3 years ago
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    Does it look right to you? I don't know if I got it right :) but there is 2 chain rule involved here

  13. DLS
    • 3 years ago
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    Isnt this what we have atm? \[\frac{dy}{dx}= -\sin(ax^{2}+bx+c) \times (2ax+b) + \cos^{3} \sqrt{ax^{2}+bx+c} \times \frac{1}{2 \sqrt{ax^{2}+bx+c}} \times 2ax+b\]

  14. DLS
    • 3 years ago
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    in the end thats 2ax+b

  15. MarcLeclair
    • 3 years ago
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    yeah thats it. Thats what you should have in my opinion :)

  16. sirm3d
    • 3 years ago
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    the chain rule applied to the second term yields \[\large \color{blue} 3 \cos ^\color{red} 2 \sqrt{ax^2+bx+c}\frac{ 1 }{ 2\sqrt{ax^2+bx+c} }\left( 2ax+b\right)\]

  17. DLS
    • 3 years ago
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    im doubtful about derivative of sin^3 as cos^3

  18. MarcLeclair
    • 3 years ago
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    nope I was right :D hehehe

  19. DLS
    • 3 years ago
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    HA! :D

  20. MarcLeclair
    • 3 years ago
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    Sirm3d. you take the derivative of the exponent of cos too?

  21. sirm3d
    • 3 years ago
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    its clearer when you write it as \[\large \left[ \sin (ax^2+bx+c)^{1/2} \right]^3\]

  22. DLS
    • 3 years ago
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    sugoi!

  23. MarcLeclair
    • 3 years ago
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    ah silly me I was thinking of sin(3x) or something

  24. DLS
    • 3 years ago
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    so u failed in ur test :P

  25. sirm3d
    • 3 years ago
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    ugh. let me retype my answer.\[ 3\left[ \sin (ax^2+bx+c)^{1/2} \right]^2\space \cos (ax^2+bx+c)^{1/2}\space \frac{ 1 }{ 2(ax^2+bx+c)^{1/2} }\left(2ax+b\right)\]

  26. DLS
    • 3 years ago
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    O___O

  27. MarcLeclair
    • 3 years ago
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    Yeah I was missing a whole term cause of it. Thanks sirm3d for correcting me!

  28. DLS
    • 3 years ago
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    okay!

  29. sirm3d
    • 3 years ago
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    that is the derivative of the second term only. the derivative of the first term provided by @MarcLeclair is correct.

  30. DLS
    • 3 years ago
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    why did we have a cos there?

  31. DLS
    • 3 years ago
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    \[\large \sin^{3}x \] suppose we have this

  32. DLS
    • 3 years ago
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    wont it be onlly 3cos^2x

  33. sirm3d
    • 3 years ago
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    3(sin x)^2 * cos x by chain rule

  34. sirm3d
    • 3 years ago
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    you should write it as (sin x)^3, then apply the chain rule

  35. DLS
    • 3 years ago
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    okay..!

  36. DLS
    • 3 years ago
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    thanks!

  37. DLS
    • 3 years ago
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    i get to learn a lot from every ques! :D

  38. sirm3d
    • 3 years ago
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    |dw:1354535975146:dw|

  39. DLS
    • 3 years ago
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    yes i got it!

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