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Math Help - Check my Derivatives =)

  1. #1
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    Check my Derivatives =)

    I hope I did these right... thanks in advance!

    g(x) = (3x-2)/(2x+1)^(1/2)

    g'(x) = (9x-1)/(2x+1)^(5/2)


    y = sin(cos(tan(x))^(1/2))

    y' = [-cos^2(tan(x))^(1/2)sin(tanx)^(1/2)sec^2(x)]/2(tan(x))^(1/2)


    these next ones dont need to be simplified

    f(x) = sin(x+x^2)

    f'(x) = cos(X^(2) + x) * (2x+1)


    f(x) = sin((cosx)/x)

    f'(x) = cos((cosx)/x) * [ (x(sinx) - cosx)/x^2]


    f(x) = sin^3(x^(2)+sinx)

    f'(x) = 3[sin(x^(2) + sinx)]^2 * cos(x^(2) + sinx) * (2x + cosx)


    f(x) = sin(xsinx) + sin(sin(x^2))

    f'(x) = cos(xsinx) * [x(cosx) + (sinx)] + cos(sin(x^(2)) * cos(x^(2)) * 2x


    heres the last one...

    Find an equation of the tangent line to the curve f(x) = sin(sinx) at the point (pi, 0)

    so i found:

    f'(x) = cos(sinx)cosx

    i plugged that into the y=mx+b as the slope of the tangent line and I got

    cos(sin(pi)) * cos(pi) =-1

    0=(-1)(pi) + b

    b= pi

    so, y= -x + pi


    Thank you so much for checking this out for me =) I will definitely try to contribute myself around here.
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  2. #2
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    Quote Originally Posted by drain View Post
    I hope I did these right... thanks in advance!

    g(x) = (3x-2)/(2x+1)^(1/2)

    g'(x) = (9x-1)/(2x+1)^(5/2)
    You can integrate it to determine if your derivative is correct.

    For the first one, I got the following:

    g(x) = (3x - 2)/(2x + 1)^(1/2)
    g'(x) = [3*(2x+1)^(1/2) - [(1/2)*(2x + 1)*2*(3x - 2)]]/2x + 1
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  3. #3
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    Quote Originally Posted by drain View Post

    y = sin(cos(tan(x))^(1/2))

    y' = [-cos^2(tan(x))^(1/2)sin(tanx)^(1/2)sec^2(x)]/2(tan(x))^(1/2)
    This one gets quite complicated, and I'll leave you to simplify it.

    y = sin(cos(tan(x))^(1/2))
    y' = cos(cos(tan(x))^(1/2))*(1/2)*[cos(tan(x))]^(-1/2)*-sin(tan(x))*sec^(x)^2
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  4. #4
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    Quote Originally Posted by AfterShock View Post
    This one gets quite complicated, and I'll leave you to simplify it.

    y = sin(cos(tan(x))^(1/2))
    y' = cos(cos(tan(x))^(1/2))*(1/2)*[cos(tan(x))]^(-1/2)*-sin(tan(x))*sec^(x)^2
    One other thing you have to realize in this prob; cos(tan(x)) is the term with the square root over it, not just tan(x).
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  5. #5
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    Quote Originally Posted by drain View Post

    f(x) = sin(x+x^2)

    f'(x) = cos(X^(2) + x) * (2x+1)
    f(x) = sin(x + x^2)
    f'(x) = cos(x + x^2)*(1 + 2x)
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  6. #6
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    Quote Originally Posted by drain View Post

    f(x) = sin((cosx)/x)

    f'(x) = cos((cosx)/x) * [ (x(sinx) - cosx)/x^2]
    f(x) = sin((cos(x))/x)

    Your derivative is almost right; you have a sign error. It should be:

    f'(x) = cos(cos(x)/x)*[(-sin(x)/x) - cos(x)/x^2]
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  7. #7
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    Quote Originally Posted by drain View Post
    f(x) = sin^3(x^(2)+sinx)

    f'(x) = 3[sin(x^(2) + sinx)]^2 * cos(x^(2) + sinx) * (2x + cosx)
    Wow, how many of these do you have. I'll do this last one for now.

    f(x) = (sin(x^(2) + sin(x)))^3

    f'(x) = 3*sin(x^2 + sin(x))^2*cos(x^2 + sin(x))*(2x + cos(x))

    Yup, it looks like your derivative is what I have.
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  8. #8
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    Quote Originally Posted by AfterShock View Post
    One other thing you have to realize in this prob; cos(tan(x)) is the term with the square root over it, not just tan(x).
    It was just the tan(x) I think I typed it it wrong =( sorry

    Edit: I think I typed it right? It was supposed to be just the tan(x) in the square root.

    Quote Originally Posted by AfterShock View Post
    f(x) = sin((cos(x))/x)

    Your derivative is almost right; you have a sign error. It should be:

    f'(x) = cos(cos(x)/x)*[(-sin(x)/x) - cos(x)/x^2]
    I had the minus sign, I mistyped again =(

    Quote Originally Posted by AfterShock View Post
    You can integrate it to determine if your derivative is correct.
    I don't know what that means or how to do that.


    THANK YOU VERY MUCH FOR YOUR TIME AND EFFORT. I really appreciate it, man. Thanks again.
    Last edited by drain; March 6th 2007 at 10:26 PM.
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