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DLS

  • 2 years ago

siny=xsin(a+y) dy/dx=?

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  1. DLS
    • 2 years ago
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    I have a small doubt here..

  2. DLS
    • 2 years ago
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    is xsin(a+y) to be solved with product rule?

  3. lgbasallote
    • 2 years ago
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    dy/dx means you take the derivative of x... since sin(a+y) does not include any x in it... no, you don't use product rule. you treat sin(a+y) as a constant coefficient

  4. DLS
    • 2 years ago
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    :o?

  5. hartnn
    • 2 years ago
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    ummm.... u need to use product rule...

  6. DLS
    • 2 years ago
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    I did..

  7. hartnn
    • 2 years ago
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    because y is the function of x

  8. DLS
    • 2 years ago
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    What I want to ask is..if I had ysin(a+y) then I wuldve used chain rule?

  9. lgbasallote
    • 2 years ago
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    you only use product rule when two facots include x for example xsin x <-- x and sinx both have x so you use product rule xsiny <--- siny doesn't have x so treat it as coefficient remember that @hartnn ?

  10. hartnn
    • 2 years ago
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    no, product again, because both are functions of x

  11. hartnn
    • 2 years ago
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    sin y doesn't have x, right bu since y is the function of x....

  12. hartnn
    • 2 years ago
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    d/dx sin y = cos y dy/dx

  13. DLS
    • 2 years ago
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    ive done that only

  14. DLS
    • 2 years ago
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    but even WFA treats it as 0

  15. hartnn
    • 2 years ago
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    WFA ?

  16. DLS
    • 2 years ago
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    wolfaram :P ALPHA

  17. DLS
    • 2 years ago
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    spelling/*

  18. hartnn
    • 2 years ago
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    no...x sin(a+y) needs product rule.

  19. DLS
    • 2 years ago
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    :/

  20. hartnn
    • 2 years ago
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    x cos(a+y) dy/dx + sin(a+y)

  21. hartnn
    • 2 years ago
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    cos y dy/dx =x cos(a+y) dy/dx + sin(a+y) isolate dy/dx

  22. hartnn
    • 2 years ago
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    your WFA http://www.wolframalpha.com/input/?i=siny%3Dxsin%28a%2By%29%2C++dy%2Fdx%3D%3F

  23. hartnn
    • 2 years ago
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    it doesn't treat it as 0.

  24. DLS
    • 2 years ago
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    oh nvm :o

  25. DLS
    • 2 years ago
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    but still not clear why product rule..its used in product of 2 functions so x and sin(a+y) x is not a function,y is

  26. hartnn
    • 2 years ago
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    x is a function of x x^1

  27. DLS
    • 2 years ago
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    doesnt make any sense :S

  28. DLS
    • 2 years ago
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    every variable is afunction of itself then

  29. hartnn
    • 2 years ago
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    you can say that. f(x) = x^n with n=1.

  30. DLS
    • 2 years ago
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    why didnt we use chain rule for sin(a+y)

  31. hartnn
    • 2 years ago
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    we actually did.

  32. DLS
    • 2 years ago
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    :o

  33. DLS
    • 2 years ago
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    sin is one function and y other right?

  34. hartnn
    • 2 years ago
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    cos (a+y) d/dx(a+y) =cos (a+y) (0+dy/dx)

  35. hartnn
    • 2 years ago
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    yes.

  36. DLS
    • 2 years ago
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    so I have to treat x&y both as function not variables but other variable like abcd as constants(0) ?

  37. DLS
    • 2 years ago
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    when differentiating wrt x

  38. RolyPoly
    • 2 years ago
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    \[\frac{d}{dx}xsin(a+y) \]\[=sin(a+y) \frac{d}{dx}(x) + x\frac{d}{dx}sin(a+y)\]\[=sin(a+y) + x[cos(a+y)\frac{d}{dx}(a+y)]\]\[=sin(a+y) + xcos(a+y)\frac{dy}{dx}\]

  39. hartnn
    • 2 years ago
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    here since y is the function of x, we treat it as different function. if a is function of x, then a is also function..... if a is independent of x, then a is constant.

  40. DLS
    • 2 years ago
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    clear!

  41. hartnn
    • 2 years ago
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    :)

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