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Math Help - need help for differentiation problem

  1. #1
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    need help for differentiation problem

    differentiate: y = x^ ½ (1 – 2x)^ 2/3 ÷ (2 – 3x)^ 3/4 (3 – 4x)^ 4/5 x^1/2 means x power 1/2 this is how i have started, but im stuck.. is this the right way of doing it?plz help me ln y = (½ ln x)(2/3 ln (1 – 2x)) – ¾ ln (2 -3x) 4/5 ln (3 -4x)
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  2. #2
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    Is this the problem:

    \frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}x^{\frac{1}{2}}}


    If so, first notice the x^{1/2} cancels.

    You could use the product and quotient rules.
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  3. #3
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    Re:differentiation problem

    no. the problem is
    [IMG]file:///C:/DOCUME%7E1/yajna10/LOCALS%7E1/Temp/moz-screenshot-2.jpg[/IMG] y = x^1/2(1-2x)^2/3 ÷ (2-3x)^3/4(3-4x)^4/5






    sorry i made a mistake..
    can i use ln first then differentiate..
    can you show me how to do it plz??
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  4. #4
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    Hello, slash!

    Differentiate: . y \;= \;\frac{x^{\frac{1}{2}}(1 - 2x)^{\frac{2}{3}}}{(2 - 3x)^{\frac{3}{4}}(3 - 4x)^{\frac{4}{5}}}
    With such an ugly function, log differentiation is probably expected.


    Take logs: . \ln(y) \;=\;\ln\left[\frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}}\right]

    . . \ln(y) \;=\;\ln\left(x^{\frac{1}{2}}\right) - \ln(1-2x)^{\frac{2}{3}} - \ln(2-3x)^{\frac{3}{4}} - \ln(3-4x)^{\frac{4}{5}}

    . . \ln(y) \;=\;\frac{1}{2}\ln(x) + \frac{2}{3}\ln(1-2x) - \frac{3}{4}\ln(2-3x) - \frac{4}{5}\ln(3-4x)


    Differentiate implicitly: . \frac{1}{y}\!\cdot\!\frac{dy}{dx}\;=\;\frac{1}{2}\  !\cdot\!\frac{1}{x} + \frac{2}{3}\!\cdot\!\frac{\text{-}2}{1-2x} - \frac{3}{4}\!\cdot\!\frac{\text{-}3}{2-3x} - \frac{4}{5}\!\cdot\!\frac{\text{-}4}{3-4x}

    . . \frac{1}{y}\cdot\frac{dy}{dx}\;=\;\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}


    Then: . \frac{dy}{dx} \;=\;y\left[\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}\right]


    Therefore: . \frac{dy}{dx} \;\;=\;\;\frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}} \cdot\left[\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}\right]

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, slash!

    With such an ugly function, log differentiation is probably expected.


    Take logs: . \ln(y) \;=\;\ln\left[\frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}}\right]

    . . \ln(y) \;=\;\ln\left(x^{\frac{1}{2}}\right) - \ln(1-2x)^{\frac{2}{3}} - \ln(2-3x)^{\frac{3}{4}} - \ln(3-4x)^{\frac{4}{5}}

    . . \ln(y) \;=\;\frac{1}{2}\ln(x) + \frac{2}{3}\ln(1-2x) - \frac{3}{4}\ln(2-3x) - \frac{4}{5}\ln(3-4x)


    Differentiate implicitly: . \frac{1}{y}\!\cdot\!\frac{dy}{dx}\;=\;\frac{1}{2}\  !\cdot\!\frac{1}{x} + \frac{2}{3}\!\cdot\!\frac{\text{-}2}{1-2x} - \frac{3}{4}\!\cdot\!\frac{\text{-}3}{2-3x} - \frac{4}{5}\!\cdot\!\frac{\text{-}4}{3-4x}

    . . \frac{1}{y}\cdot\frac{dy}{dx}\;=\;\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}


    Then: . \frac{dy}{dx} \;=\;y\left[\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}\right]


    Therefore: . \frac{dy}{dx} \;\;=\;\;\frac{x^{\frac{1}{2}}(1-2x)^{\frac{2}{3}}}{(2-3x)^{\frac{3}{4}}(3-4x)^{\frac{4}{5}}} \cdot\left[\frac{1}{2x} - \frac{4}{3(1-2x)} + \frac{9}{4(2-3x)} + \frac{16}{5(3-4x)}\right]


    can you explain me the first line of your answer plz..why it is ln x^1/2 - ln (1-2x)^2/3 rather than (ln x^1/2)X(ln (1-2x)^2/3)
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  6. #6
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    Log. of a product

    \ln (a\cdot b)=\ln a+\ln b,\,\forall a,b\in\mathbb R^+
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