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Math Help - Finding derivative involving large powers.

  1. #1
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    Finding derivative involving large powers.

    <a href="Math.jpg picture by Calamanation - Photobucket" target="_blank"><img src="http://i971.photobucket.com/albums/a...ation/Math.jpg" border="0" alt="Photobucket"></a>



    Hopefully one of those worked, just stuck on how to work question (2b) out, can't see a solution beyond something involving dominant powers. Thanks.
    Last edited by Calamanation; January 14th 2010 at 07:03 PM. Reason: Specify which question
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    Which question in particular? What work have you done thus far?
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    Question 2b involving calculating f'(x), I haven't been able to do much because all my research hasn't come across a question like that.
    I appreciate it appears that I'm just looking for someone to answer it for me but studying for an exam has led me to asking for help.
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    It seems pretty straight forward, use the chain rule as the question is in the form y= u^{1\over 2} where

     <br />
u= \frac{(\dots)^{12}(\dots)^{23}}{(\dots)^{10}}<br />
Just being lazy here!

    Find \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}

    \frac{du}{dx} can be found using the quotient rule

    Does this help?
    Last edited by pickslides; January 14th 2010 at 10:21 PM. Reason: *sp
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    Apply \ln on both sides and derive (remember (\ln (y) )'=\frac{dy}{y} and your first exercise):

    \frac{f'(x)}{f(x)} = \frac{6(\cos (x) -\sin (x))}{4+\cos (x) + \sin (x)}  + \frac{46x}{1+2x^2} -\frac{5(3x^2+4x^3)}{1+x^3+x^4}

    Now just isolate f'.
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    Quote Originally Posted by pickslides View Post
    It seems pretty straight forward, use the chain rule as the question is in the form y= u^{1\over 2} where

     <br />
u= \frac{(\dots)^{12}(\dots)^{23}}{(\dots)^{10}}<br />
Just being lazy here!

    Find \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}

    \frac{du}{dx} can be found using the quotient rule

    Does this help?
    Helps heaps, thanks for that, very much appreciated
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  7. #7
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    Quote Originally Posted by Calamanation View Post
    <a href="Math.jpg picture by Calamanation - Photobucket" target="_blank"><img src="http://i971.photobucket.com/albums/a...ation/Math.jpg" border="0" alt="Photobucket"></a>



    Hopefully one of those worked, just stuck on how to work question (2b) out, can't see a solution beyond something involving dominant powers. Thanks.
    2. b) is probably easiest if you realise

    \sqrt{\frac{(4 + \cos{x} + \sin{x})^{12}(1 + 2x^2)^{23}}{(1 + x^3 + x^4)^{10}}} = \frac{(4 + \cos{x} + \sin{x})^6(1 + 2x^2)^{\frac{23}{2}}}{(1 + x^3 + x^4)^{5}}.

    Now you don't need to use chain rule inside chain rule.


    2. c) \frac{dy}{dx} = x^2 y

    \frac{1}{y}\,\frac{dy}{dx} = x^2

    \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = \int{x^2\,dx}

    \int{\frac{1}{y}\,dy} = \frac{1}{3}x^3 + C_1

    \ln{|y|} + C_2 = \frac{1}{3}x^3 + C_1

    \ln{|y|} = \frac{1}{3}x^3 + C, where C = C_1 - C_2

    |y| = e^{\frac{1}{3}x^3 + C}

    |y| = e^Ce^{\frac{1}{3}x^3}

    y = \pm e^C e^{\frac{1}{3}x^3}

    y = Ae^{\frac{1}{3}x^3}, where A = \pm e^C.

    You are told y(0) = 1, so

    1 = Ae^{\frac{1}{3}(0)^3}

    1 = Ae^0

    1 = A.


    So y = e^{\frac{1}{3}x^3}.


    Can you do the rest of that question?
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    Also, to prove

    \ln{(XY)} = \ln{X} + \ln{Y}

    we look at the properties of exponentials.


    We know that

    a^x\cdot a^y = a^{x + y}.

    So taking the logarithm of that base to both sides, we get

    \log_a{(a^x \cdot a^y)} = \log_a{(a^{x + y})}

    \log_a{(a^x \cdot a^y)} = x + y.


    And since x = \log_a{a^x} and y = \log_a{a^y}, we find

    \log_a{(a^x \cdot a^y)} = \log_a{a^x} + \log_a{a^y}.


    Using e as your base and X = a^x and Y = a^y completes your proof.
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