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Math Help - Derivatives derivatives...

  1. #1
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    Derivatives derivatives...

    man i am just getting stump on every question of my practice exam...
    heres where im getting lost

    Complete the following

    du/dv, where u=y^2 and v=y+3y^3 where do i start? take the derivative of each?


    dy/dx where y=f(1-2x) and f ' (x)= sec (x)


    dy/dx at the point (1,2), where y=y^2-2x^3
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  2. #2
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    Quote Originally Posted by cyberdx16 View Post
    man i am just getting stump on every question of my practice exam...
    heres where im getting lost

    Complete the following

    du/dv, where u=y^2 and v=y+3y^3 where do i start? take the derivative of each?
    Paramentric curve.

    Find the derivative of the first,
    u'=2y
    Now the derivative of the second,
    v'=1+9y^2
    Then their ratio is the derivative,
    u'/v'=2y/(1+9y^2)

    dy/dx where y=f(1-2x) and f ' (x)= sec (x)
    Do not understand.

    dy/dx at the point (1,2), where y=y^2-2x^3
    Implictly:
    y'=2yy'-6x^2
    Substitute (1,2)
    y'=2(2)y'-6(1)^2
    Thus,
    y'=4y'-6
    Thus,
    -3y'=-6
    Thus,
    y'=2
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Paramentric curve.

    Find the derivative of the first,
    u'=2y
    Now the derivative of the second,
    v'=1+9y^2
    Then their ratio is the derivative,
    u'/v'=2y/(1+9y^2)


    Do not understand.



    Implictly:
    y'=2yy'-6x^2
    Substitute (1,2)
    y'=2(2)y'-6(1)^2
    Thus,
    y'=4y'-6
    Thus,
    -3y'=-6
    Thus,
    y'=2

    how would u go about and find the second derivative? as well
    how do u get y'=2yy'-6x^2, is because when dy/dx, which means to find y as the function of x? thx!
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  4. #4
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    Quote Originally Posted by cyberdx16 View Post
    man i am just getting stump on every question of my practice exam...
    heres where im getting lost

    Complete the following

    du/dv, where u=y^2 and v=y+3y^3 where do i start? take the derivative of each?
    du/dv=du/dy dy/dv

    Also when v is a well enough behaved function of y dy/dv=1/(dv/dy).

    Now du/dy=2y, and dv/dy=1+9y^2, so:

    du/dv=2y/(1+9y^2)

    Now we can if we wish rewrite the right hand side in terma of u and v.

    du/dv=2y/(1+9u),

    and as

    v=y+3y^3=y(1+3y^2)=y(1+3u)

    we have:

    y=v/(1+3u),

    hence:

    du/dv=2v/[(1+3u)(1+9u)]
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  5. #5
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    Quote Originally Posted by cyberdx16 View Post

    dy/dx where y=f(1-2x) and f ' (x)= sec (x)
    You need the chain rule here:

    d/dx [f(g(x))] = f'(g(x))g'(x)

    So here we have g(x)=1-2x, so g'=-2 so:

    dy/dx = f'(1-2x)(-2) = -2 sec(1-2x)

    RonL
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