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Math Help - Derivatives and Tangent lines

  1. #1
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    Derivatives and Tangent lines

    As always, I have a very loose definition of what these things mean, so going slow would be appreciated.

    1) g(x) = sqrt(9 + 4x)

    2) 9cos(x) + (5/7)cot(x)

    3) y = (6 + sin(x)) / (6x + cos(x)) Find the derivative

    4) sec(x) Find the second derivative fprime(pi/3)

    5) y = e^(9e^x))

    6) y = sqrt( (1-x) / (1 + x))

    I really don't have a clue what I'm doing here.
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  2. #2
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    Quote Originally Posted by CFem View Post
    As always, I have a very loose definition of what these things mean, so going slow would be appreciated.

    1) g(x) = sqrt(9 + 4x)

    chain rule

    2) 9cos(x) + (5/7)cot(x)

    straight-forward derivative of two trig functions

    3) y = (6 + sin(x)) / (6x + cos(x)) Find the derivative

    quotient rule

    4) sec(x) Find the second derivative fprime(pi/3)

    derivative of the derivative ... 2nd derivative will involve the product rule.

    5) y = e^(9e^x))

    chain rule again

    6) y = sqrt( (1-x) / (1 + x))

    chain rule in conjunction with the quotient rule ... unless you know logarithmic differentiation.

    I really don't have a clue what I'm doing here.

    then get one ... go to the link.
    Pauls Online Notes : Calculus I - Derivatives
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  3. #3
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    I think I've messed up the Chain Rule somehow.

    Working with number 1, I ended up with 1/(18 + 8x)^(1/2)

    I cut the function up into sqrt(x) and 9 + 4x
    Derivatives being (1/2)x^(1/2) and 4 respectively

    => ((1/2)( 9 + 4x)^(1/2)) * 4

    Where did I go wrong?
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  4. #4
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    Quote Originally Posted by CFem View Post
    I think I've messed up the Chain Rule somehow.

    Working with number 1, I ended up with 1/(18 + 8x)^(1/2)

    I cut the function up into sqrt(x) and 9 + 4x
    Derivatives being (1/2)x^(1/2) and 4 respectively

    => ((1/2)( 9 + 4x)^(1/2)) * 4

    Where did I go wrong?
    you cannot "split" two terms inside a radical ... \sqrt{a+b} \ne \sqrt{a} + \sqrt{b} ... review your algebra.


    y = \sqrt{9+4x}

    y = (9+4x)^{\frac{1}{2}}

    \frac{dy}{dx} = \frac{1}{2}(9+4x)^{-\frac{1}{2}} \cdot (4)<br />

    \frac{dy}{dx} = 2(9+4x)^{-\frac{1}{2}} = \frac{2}{\sqrt{9+4x}}
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  5. #5
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    That's what I had, I just don't think it translated well over text.

    But thanks. I lost a negative sign it looks like.
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    Sorry if double posting isn't allowed, but I didn't think it would be necessary to make another topic.

    On number 3:

    f(x) = 6 + sin(x)
    f'(x) = cos(x)

    g(x) = 6x + cos(x)
    g'(x) = 6 - sin(x)

    (6x + cos(x))(cos(x)) - (6 + sin(x))(6 - sin(x))
    -------------------------------------
    (6 + cos(x))^2

    6xcos(x) + cos(x)^2 -36 + sin(x)^2
    ---------------------------------
    36 + 12cos(x) + cos(x)^2

    =

    6xcosx + sin(x)^2
    -----------------
    12cos(x)

    I have a feeling I made another simple mistake or I forgot to simplify something...
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  7. #7
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    Quote Originally Posted by CFem View Post
    Sorry if double posting isn't allowed, but I didn't think it would be necessary to make another topic.

    On number 3:

    f(x) = 6 + sin(x)
    f'(x) = cos(x)

    g(x) = 6x + cos(x)
    g'(x) = 6 - sin(x)

    (6x + cos(x))(cos(x)) - (6 + sin(x))(6 - sin(x))
    -------------------------------------
    (6 + cos(x))^2

    6xcos(x) + cos(x)^2 -36 + sin(x)^2
    ---------------------------------
    36 + 12cos(x) + cos(x)^2

    ... you cannot "cancel" the cos^2(x)'s , that is an algebra no-no.
    ... note (in the numerator) that cos^2(x) + sin^2(x) = 1
    ...
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  8. #8
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    Clearly I am not understanding this
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  9. #9
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    Quote Originally Posted by CFem View Post
    Clearly I am not understanding this
    your calculus on the last question was fine ... it's your algebra that is hurting you.
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    6xcos(x) + 1 - 1
    ----------------
    12cos(x) + cos(x)^2

    It comes back to Trig more than anything.

    Not sure where to go from here.
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  11. #11
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    Quote Originally Posted by CFem View Post
    On number 3:

    f(x) = 6 + sin(x)
    f'(x) = cos(x)

    g(x) = 6x + cos(x)
    g'(x) = 6 - sin(x)

    (6x + cos(x))(cos(x)) - (6 + sin(x))(6 - sin(x))
    -------------------------------------
    (6x + cos(x))^2 ... leave this alone, squaring it does little to help.


    6xcos(x) + cos(x)^2 - 36 + sin(x)^2
    ---------------------------------
    [6x + cos(x)]^2


    6xcosx + 1 - 36
    -----------------
    [6x + cos(x)]^2


    6xcosx - 35
    ---------------
    [6x + cos(x)]^2
    you're done.
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  12. #12
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    Our homework gets checked by some sort of automated system, which spat that out.

    Thanks anyway.
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  13. #13
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    3) F(x) = e^(9e^x))

    f(x) = e^x
    f'(x) = e^x

    g(x) = 9e^x
    g'(x) = 9(d/dx)e^x = 9e^x

    F'(x) = f'(g(x))(g'(x))

    => e^(9^e^x) * 9e^x
    => 9e^(9xe^x)

    More Algebra problems?

    something) Find the tangent line of x^4 + 8e^x

    I ended up using L'Hospital's Rule and got the slope of the tangent line to = 8 (0 without the rule) and ended up with y = -8(x-1), but I lost my work on this one. Ideas?
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  14. #14
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    the general derivative for y = e^u, where u is a function of x ...

    \frac{d}{dx} [e^u] = e^u \cdot \frac{du}{dx}



    for f(x) = e^{9e^x} , the u is 9e^x ... \frac{du}{dx} = 9e^x

    so ... f'(x) = e^{9e^x} \cdot 9e^x = 9e^{9e^x} \cdot e^x = 9e^{9e^x+x}
    Last edited by skeeter; September 27th 2009 at 05:00 PM. Reason: fix exponent
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  15. #15
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    Quote Originally Posted by CFem View Post
    something) Find the tangent line of x^4 + 8e^x

    I ended up using L'Hospital's Rule and got the slope of the tangent line to = 8 (0 without the rule) and ended up with y = -8(x-1), but I lost my work on this one. Ideas?
    tangent line at what point?

    L'Hopital is for limits ... ???
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