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Math Help - derivative of y=sec(xy^3) ?!

  1. #1
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    derivative of y=sec(xy^3) ?!

    derivative of y=sec(xy^3) ?!!

    y= sec x + sec y^3 ??

    anyone help me out please
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  2. #2
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    Quote Originally Posted by haebinpark View Post
    derivative of y=sec(xy^3) ?!!

    y= sec x + sec y^3 ?? no
    \frac{dy}{dx} = \sec(xy^3)\tan(xy^3)[x \cdot 3y^2 \frac{dy}{dx} + y^3]

    solve for \frac{dy}{dx}
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  3. #3
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    Quote Originally Posted by haebinpark View Post
    derivative of y=sec(xy^3) ?!!

    y= sec x + sec y^3 ??

    anyone help me out please
    \sec(xy^3)-y=0

    Use implicit differentiation.

    Note that the derivative of \sec u is \sec u\tan u.
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    Quote Originally Posted by haebinpark View Post
    derivative of y=sec(xy^3) ?!!

    y= sec x + sec y^3 ??

    anyone help me out please
    This one you have to use the chain rule with implicit differentiation.

    y=sec(u) so \frac{dy}{dx}=sec(u)tan(u)\frac{du}{dx} where u=xy^3

    The derivative is therefore:

    (y^3+x3y^2y')sec(xy^3)tan(xy^3)
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  5. #5
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    what do you mean by (y^3 + x*3*y^2*y') ?

    i ended up getting answer : [sec(xy^3)][tan(xy^3)][y^3 + 3xy^2]

    is that correct?
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    Quote Originally Posted by haebinpark View Post
    what do you mean by (y^3 + x*3*y^2*y') ?

    i ended up getting answer : [sec(xy^3)][tan(xy^3)][y^3 + 3xy^2]

    is that correct?
    You're not applying the chain rule to the second term in:

    y^3 + x(y^3)'

    (y^3)' =3y^2y'

    So the derivative of xy^3 is y^3+3xy^2y'
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  7. #7
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    i can't leave y' in the answer, can i?
    if not, what is the next step?


    edited:

    ohhhhh

    y' of y=sec(xy^3) again???????
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  8. #8
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    [sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']
    = [sec(xy^3)][tan(xy^3)][y^3 + 3xy^2(sec(xy^3))(tan(xy^3))]

    would be the answer? or can be more simple?
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  9. #9
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    Quote Originally Posted by haebinpark View Post
    [sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']
    = [sec(xy^3)][tan(xy^3)][y^3 + 3xy^2(sec(xy^3))(tan(xy^3))]

    would be the answer? or can be more simple?
    The part that I highlighted in red is correct. Remember that we used implicit differentiation. Suppose I want to differentiate the equation:

    y=3x^2+y^3

    I have to use implicit differentiation:

    y'=6x+3y^2y'

    Notice that y' is on both sides of the equation. You can often solve for y'.

    y'+-3y^2y'=6x

    y'(1-3y^2)=6x

    y'=\frac{6x}{1-3y^2}

    With your equation, we can do the same thing:

    y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']

    y'-3xy^2y'sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)

    y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)

    y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}
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  10. #10
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    Quote Originally Posted by adkinsjr View Post
    The part that I highlighted in red is correct. Remember that we used implicit differentiation. Suppose I want to differentiate the equation:

    y=3x^2+y^3

    I have to use implicit differentiation:

    y'=6x+3y^2y'

    Notice that y' is on both sides of the equation. You can often solve for y'.

    y'+-3y^2y'=6x

    y'(1-3y^2)=6x

    y'=\frac{6x}{1-3y^2}

    With your equation, we can do the same thing:

    y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']

    sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)" alt="y'-3xy^2y'sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)" />

    y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)

    y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}

    the line i highlighted, where did it come from?
    shouldn't it be just 3xy^2 y' ?
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  11. #11
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    Lightbulb

    Quote Originally Posted by adkinsjr View Post
    The part that I highlighted in red is correct. Remember that we used implicit differentiation. Suppose I want to differentiate the equation:

    y=3x^2+y^3

    I have to use implicit differentiation:

    y'=6x+3y^2y'

    Notice that y' is on both sides of the equation. You can often solve for y'.

    y'+-3y^2y'=6x

    y'(1-3y^2)=6x

    y'=\frac{6x}{1-3y^2}

    With your equation, we can do the same thing:

    y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']

    sec(xy^3)tan(xy^3) =y^3sec(xy^3)tan(xy^3)" alt="y'-3xy^2y' sec(xy^3)tan(xy^3) =y^3sec(xy^3)tan(xy^3)" />

    y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)

    y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}

    the line i highlighted, where did it come from?
    shouldn't it be just 3xy^2 y' ?
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  12. #12
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    i can't highlight for some reason

    i meant
    you wrote
    y' - 3xy^2y'sec(xy^3)tan(xy^3) = blah

    but shouldn't it be just 3xy^2 y' instead of 3xy^2y'sec(xy^3)tan(xy^3)?
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  13. #13
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    Quote Originally Posted by haebinpark View Post
    i can't highlight for some reason

    i meant
    you wrote
    y' - 3xy^2y'sec(xy^3)tan(xy^3) = blah

    but shouldn't it be just 3xy^2 y' instead of 3xy^2y'sec(xy^3)tan(xy^3)?
    You have to EXPAND the right side of this equation.

    y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']

    y'=y^3sec(xy^3)tan(xy^3)+3xy^2y'sec(xy^3)tan(xy^3)

    y'-3xy^2y'sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)

    y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)

    y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}
    Last edited by adkinsjr; October 29th 2009 at 02:48 PM.
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  14. #14
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    Quote Originally Posted by adkinsjr View Post
    You have to factor the right side of this equation.

    y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y']

    3xy^2y'sec(xy^3)tan(xy^3)

    y'-3xy^2y'sec(xy^3)tan(xy^3)=y^3sec(xy^3)tan(xy^3)

    y'[1-3xy^2sec(xy^3)tan(xy^3)]=y^3sec(xy^3)tan(xy^3)

    y'=\frac{y^3sec(xy^3)tan(xy^3)}{1-3xy^2sec(xy^3)tan(xy^3)}
    i still don't see where sec(xy^3)tan(xy^3) came from :s (after 3xy^2y')
    3xy^2y'sec(xy^3)tan(xy^3)
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  15. #15
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    Quote Originally Posted by haebinpark View Post
    i still don't see where sec(xy^3)tan(xy^3) came from :s (after 3xy^2y')
    3xy^2y'sec(xy^3)tan(xy^3)
    When you expand the right side of the equation: y'=sec(xy^3)][tan(xy^3)][y^3 + 3xy^2y'] you use the distributive property of algebra. To make this more obvious, let A=sec(xy^3)tan(xy^3). So the equation becomes:

    y'=A(y^3+3xy^2y')=Ay^3+A3xy^2y'

    Then I subtracted the A3xy^2y' from both sides and factored the left side to isolate y'.
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