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Math Help - Differentiation Question

  1. #1
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    Differentiation Question

    So I've been trying to differentiate this one problem, and then solve when the derivative is equal to 2. The equation in question is sqrt[(5x+1)/(4x^2+1)

    Now for my answer before plugging in two I got .5((((20x^2+5)-(40x^2+8x))/(16x^4+8x^2+1))^(-1/2)

    however this is obviously not a correct calculation - what was I doing wrong?
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  2. #2
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    This site helped me alot.

    http://www.mathhelpforum.com/math-he...step-step.html
    solving derivatives step-by-step

    Just type in sqrt[(5x+1)/(4x^2+1)].
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Pikeman85 View Post
    So I've been trying to differentiate this one problem, and then solve when the derivative is equal to 2. The equation in question is sqrt[(5x+1)/(4x^2+1)

    Now for my answer before plugging in two I got .5((((20x^2+5)-(40x^2+8x))/(16x^4+8x^2+1))^(-1/2)

    however this is obviously not a correct calculation - what was I doing wrong?
    y=\sqrt{\frac{5x+1}{4x^2+1}}\Rightarrow{y^2=\frac{  5x+1}{4x^2+1}}

    Differentiating we get

    2y\cdot{y'}=\frac{-20x^2-8x+5}{(4x^2+1)^2}

    So remembering that y=\sqrt{\frac{5x+1}{4x^2+1}}

    we see the answer is

    y'=\frac{-20x^2-8x+5}{2\sqrt{\frac{5x+1}{4x^2+1}}(4x^2+1)^2}


    so y'(2)=\frac{-91\sqrt{187}}{6350}\approx{-.195723}
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Pikeman85 View Post
    So I've been trying to differentiate this one problem, and then solve when the derivative is equal to 2. The equation in question is sqrt[(5x+1)/(4x^2+1)

    Now for my answer before plugging in two I got .5((((20x^2+5)-(40x^2+8x))/(16x^4+8x^2+1))^(-1/2)

    however this is obviously not a correct calculation - what was I doing wrong?
    y=\sqrt{\frac{5x+1}{4x^2+1}}

    y^{/}=\frac{1}{2}\left(\frac{5x+1}{4x^2+1}\right)^{-\frac{1}{2}}\cdot \left(\frac{(8x)(5x+1)-(4x^2+1)(5)}{(4x^2+1)^2}\right)

    Try taking it from here.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    y=\sqrt{\frac{5x+1}{4x^2+1}}

    y^{/}=\frac{1}{2}\left(\frac{5x+1}{4x^2+1}\right)^{-\frac{1}{2}}\cdot \left(\frac{(8x)(5x+1)-(4x^2+1)(5)}{(4x^2+1)^2}\right)

    Try taking it from here.
    Why do you put

    y^{/} instead of y'


    I think the latter looks nicer...not that it matters...I am just curious
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  6. #6
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    RE:diff

    I punched you question into matlab, the derivative of your function

    f(x) = sqrt(((5*x+1)/(4*x^2+1)))

    is

    f '(x) = 1/2/((5*x+1)/(4*x^2+1))^(1/2)*(5/(4*x^2+1)-8*(5*x+1)/(4*x^2+1)^2*x)

    Im certain its 100% correct since a computer calculated it, if I'm wrong, let me know
    Last edited by martinr; May 22nd 2008 at 04:28 PM. Reason: misread equation first time
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Why do you put

    y^{/} instead of y'


    I think the latter looks nicer...not that it matters...I am just curious
    Its just a habbit. I do this a lot when I use mathtype, for it looks better that (').
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by martinr View Post
    I punched you question into matlab, the derivative of your function

    f(x) = sqrt(((5*x+1)/(4*x^2+1)))

    is

    f '(x) = 1/2/((5*x+1)/(4*x^2+1))^(1/2)*(5/(4*x^2+1)-8*(5*x+1)/(4*x^2+1)^2*x)

    Im certain its 100% correct since a computer calculated it, if I'm wrong, let me know
    Mathematica gave me this:



    I assume its the same??
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  9. #9
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    Hrm, here is another problem, this one is VERY difficult for me to solve.

    [IMG]file:///D:/DOCUME%7E1/ANDREW%7E2.MAI/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG] Let F(x)= f(x^{5}) and G(x)=(f(x))^{5} . You also know that a^{4}=11, f(a)=3,f'(a)=10, f'(a^{5})=5.


    Find F'(a)= and G'(a)=

    Where do I even start with a problem like this one?
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Pikeman85 View Post
    Hrm, here is another problem, this one is VERY difficult for me to solve.

    [IMG]file:///D:/DOCUME%7E1/ANDREW%7E2.MAI/LOCALS%7E1/Temp/moz-screenshot.jpg[/IMG] Let F(x)= f(x^{5}) and G(x)=(f(x))^{5} . You also know that a^{4}=11, f(a)=3,f'(a)=10, f'(a^{5})=5.


    Find F'(a)= and G'(a)=

    Where do I even start with a problem like this one?
    F(x)=f(x^5)

    differentiating using the chain rule we get

    F'(x)=f'(x^5)\cdot{5x^4}

    so F'(a)=f'(a^5)\cdot{5a^4}

    and since they give us f(a^5)=5 and a^4=11

    we have F'(5)=5\cdot{5}\cdot{11}=275
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  11. #11
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    I'd generally prefer it if people gave me more hints than answers. Oh work through part of a problem maybe, but eh. I want to figure this stuff out myself so that I actually KNOW it.

    Now this one - this one is difficult for me (and thanks for everyone that has helped me this time and the time before! I'm actually getting some of the differentiation stuff quite a bit better now - still need to work on the Chain Rule, and a couple other things, but I have many of the others)

    Ok so here's the problem I have now:

    f(x) =
    { x^2 + 6x + 13 X (less than or equal to sign) 2
    { ax + b X (greater than) 2


    Find a and b such that the function is differentiable everywhere. So I'm looking for continuity, and I know that at 2, the top equation = 28.

    How do I go about solving this?
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Pikeman85 View Post
    I'd generally prefer it if people gave me more hints than answers. Oh work through part of a problem maybe, but eh. I want to figure this stuff out myself so that I actually KNOW it.

    Now this one - this one is difficult for me (and thanks for everyone that has helped me this time and the time before! I'm actually getting some of the differentiation stuff quite a bit better now - still need to work on the Chain Rule, and a couple other things, but I have many of the others)

    Ok so here's the problem I have now:

    f(x) =
    { x^2 + 6x + 13 X (less than or equal to sign) 2
    { ax + b X (greater than) 2


    Find a and b such that the function is differentiable everywhere. So I'm looking for continuity, and I know that at 2, the top equation = 28.

    How do I go about solving this?
    You dont have to CAPITALIZE things ...I understand


    Ok so you want a hint but not an answer

    Here is your hint

    if f(x) is differentiable at c then

    \lim_{x\to{c^{+}}}f(x)=\lim_{x\to{c^{-}}}f(x)

    and \lim_{x\to{c^{-}}}f'(x)=\lim_{x\to{c^{+}}}f'(x)
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  13. #13
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    Yes, I know that the two points have to touch, the problem is I am not sure how I am supposed to calculate which two points do touch
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Pikeman85 View Post
    Yes, I know that the two points have to touch, the problem is I am not sure how I am supposed to calculate which two points do touch
    Ok, so what you must do is this for it to be differentiable at the problem spot x=2

    f(x) to the left of two must equal f(x) from the right of two

    and f'(x) from the left of two must equal f'(x) from the right of two

    so now you have to equations with two unknowns. Solve for the unknowns
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  15. #15
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    Still not getting it. I'm very dense about a lot of the simpler bits. I've got most of the differentiation ones done, but this one eludes me
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